训练赛题解
突然想到好久以前做完這份題目沒寫題解。蠻來寫寫吧。很多細節已經忘記了。。
第一題
很簡單的字符串比對是否b包含a。不包含就報NO，包含就YES。。坑爹的第一次！！。把strlen放在了for循環里面。。就超時了。。超時了。。
注意：for里面的條件每次也會重新計算。
A - All in All Time Limit:1000MS???? Memory Limit:30000KB???? 64bit IO Format:%I64d & %I64u Submit Status Practice POJ 1936Description
You have devised a new encryption technique which encodes a message by inserting between its characters randomly generated strings in a clever way. Because of pending patent issues we will not discuss in detail how the strings are generated and inserted into the original message. To validate your method, however, it is necessary to write a program that checks if the message is really encoded in the final string.Given two strings s and t, you have to decide whether s is a subsequence of t, i.e. if you can remove characters from t such that the concatenation of the remaining characters is s.
Input
The input contains several testcases. Each is specified by two strings s, t of alphanumeric ASCII characters separated by whitespace.The length of s and t will no more than 100000.Output
For each test case output "Yes", if s is a subsequence of t,otherwise output "No".Sample Input
sequence subsequence person compression VERDI vivaVittorioEmanueleReDiItalia caseDoesMatter CaseDoesMatterSample Output
Yes No Yes No下附代碼 1 #include<iostream> 2 #include<stdio.h> 3 #include<string.h> 4 #include<cmath> 5 #define INF 1000000 6 char s[100005],t[100005]; 7 int main() 8 { 9 int i=0,j=0,s_len,t_len; 10 while(scanf("%s%s",s,t)!=EOF) 11 { 12 i=0;j=0; 13 s_len=strlen(s); 14 t_len=strlen(t); 15 for(i;i<s_len&&j<t_len;) 16 if(s[i]==t[j]) 17 {i++;j++;} 18 else 19 j++; 20 if(i==strlen(s)) 21 printf("Yes\n"); 22 else 23 printf("No\n"); 24 } 25 return 0; 26 }
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第二題
找規律。。其實也可以還原。但是找到規律的話也是可以做的。就是寫代碼的時候要認真點。思路是相鄰做差，得到一個新的數組，新數組有數字，說明肯定有差括號，是1，然后這個新數組如果有一串0，說明原來的數字都是相等的，則第一位0肯定是1，但是接下來的0，就要繼續往前找對應的括號，然后增加數字，具體的已經忘記了。但是就是找規律。這題1A。
B - Parencodings Time Limit:1000MS???? Memory Limit:10000KB???? 64bit IO Format:%I64d & %I64u Submit Status Practice POJ 1068Description
Let S = s1 s2...s2n be a well-formed string of parentheses. S can be encoded in two different ways:q By an integer sequence P = p1 p2...pn where pi is the number of left parentheses before the ith right parenthesis in S (P-sequence).
q By an integer sequence W = w1 w2...wn where for each right parenthesis, say a in S, we associate an integer which is the number of right parentheses counting from the matched left parenthesis of a up to a. (W-sequence).
Following is an example of the above encodings:
S (((()()())))
P-sequence 4 5 6666
W-sequence 1 1 1456
Write a program to convert P-sequence of a well-formed string to the W-sequence of the same string.
Input
The first line of the input contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case is an integer n (1 <= n <= 20), and the second line is the P-sequence of a well-formed string. It contains n positive integers, separated with blanks, representing the P-sequence.Output
The output file consists of exactly t lines corresponding to test cases. For each test case, the output line should contain n integers describing the W-sequence of the string corresponding to its given P-sequence.Sample Input
2 6 4 5 6 6 6 6 9 4 6 6 6 6 8 9 9 9Sample Output
1 1 1 4 5 6 1 1 2 4 5 1 1 3 9下附代碼： 1 #include<iostream> 2 #include<stdio.h> 3 #include<string.h> 4 #include<cmath> 5 #include<algorithm> 6 #include<queue> 7 #define INF 1000005 8 using namespace std; 9 queue<int>q; 10 int s[1000005]={0},mark[1000005]; 11 int main() 12 { 13 int n,first,late,i,m,ans,t; 14 scanf("%d",&n); 15 while(n--) 16 { 17 for(i=0;i<30;i++) 18 {s[i]=0;mark[i]=0;} 19 scanf("%d",&m); 20 first=0; 21 for(i=0;i<m;i++) 22 { 23 scanf("%d",&late); 24 s[i]=late-first; 25 first=late; 26 } 27 for(i=0;i<m;i++) 28 { 29 if(s[i]) 30 { 31 if(i!=m-1) 32 printf("1 "); 33 else 34 printf("1\n"); 35 } 36 else 37 { 38 t=i;ans=0; 39 while(!s[t]||mark[t]==s[t]) 40 { 41 ans+=mark[t]; 42 t--; 43 } 44 while(mark[t]+1==s[t]) 45 { 46 mark[t]++; 47 ans+=s[t]; 48 t--; 49 while(!s[t]||mark[t]==s[t]) 50 { 51 ans+=mark[t]; 52 t--; 53 } 54 } 55 mark[t]++; 56 if(i!=m-1) 57 printf("%d ",ans+mark[t]+1); 58 else 59 printf("%d\n",ans+mark[t]+1); 60 } 61 } 62 } 63 return 0; 64 }
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第三題
一個二叉樹的前序中序序列，還原后序序列，用到了遞歸分治的思想，主要要分清楚兩個序列的頭和尾在哪里，然后可以通過找根找到中序序列中間的根，從而繼續分開這兩個序列。關鍵代碼是下面這個
if(k-cc)turn(aa+1,k-cc+aa,cc,k-1); if(dd-k) turn(aa+k-cc+1,bb,k+1,dd);錯了兩次，算是一道考細節的遞歸。
下附代碼：
1 #include<iostream> 2 #include<stdio.h> 3 #include<string.h> 4 #include<cmath> 5 #include<algorithm> 6 #include<queue> 7 #define INF 1000000 8 using namespace std; 9 char a[50],b[50],c[50]; 10 void turn(int aa,int bb,int cc,int dd)//n是前序m是中序 11 { 12 int k=0; 13 while(a[aa]!=b[k]) 14 { 15 k++; 16 }//找根 17 if(k-cc) 18 turn(aa+1,k-cc+aa,cc,k-1); 19 if(dd-k) 20 turn(aa+k-cc+1,bb,k+1,dd); 21 printf("%c",a[aa]); 22 } 23 int main() 24 { 25 int i,m,n,ans,t,bg,j,num,k,h=0,sum=0; 26 while(scanf("%s%s",a,b)!=EOF) 27 { 28 t=strlen(a); 29 turn(0,t-1,0,t-1); 30 for(i=1;i<=t;i++) 31 printf("%c",c[i]); 32 printf("\n"); 33 } 34 return 0; 35 }
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第四題：
BFS就行。用隊列來存，注意一下具體細節。
D - Catch That Cow Time Limit:2000MS???? Memory Limit:65536KB???? 64bit IO Format:%I64d & %I64u Submit Status Practice POJ 3278Description
Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.
* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.
If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?
Input
Line 1: Two space-separated integers: N and KOutput
Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.Sample Input
5 17Sample Output
4Hint
The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.下附代碼：
1 #include<iostream> 2 #include<stdio.h> 3 #include<string.h> 4 #include<cmath> 5 #include<algorithm> 6 #include<queue> 7 #define INF 1000000 8 using namespace std; 9 queue<int>q; 10 int vis[1000005]={0}; 11 int main() 12 { 13 int a,b,cot,ans,t,mark=0,noww; 14 scanf("%d%d",&a,&b); 15 if(a==b) 16 printf("0\n"); 17 else 18 { 19 if(a+1==b||a-1==b||a*2==b) 20 printf("1\n"); 21 else 22 { 23 q.push(a); 24 vis[a]=1; 25 while(!q.empty()) 26 { 27 ans=q.front(); 28 q.pop(); 29 if(ans-1>=0) 30 { 31 noww=ans-1; 32 if(!vis[noww]) 33 {vis[noww]=vis[ans]+1;q.push(noww);} 34 if(noww==b) 35 {printf("%d\n",vis[noww]-1);break;} 36 } 37 if(ans+1<INF) 38 { 39 noww=ans+1; 40 if(!vis[noww]) 41 {vis[noww]=vis[ans]+1;q.push(noww);} 42 if(noww==b) 43 {printf("%d\n",vis[noww]-1);break;} 44 } 45 if(ans*2<INF) 46 { 47 noww=ans*2; 48 if(!vis[noww]) 49 {vis[noww]=vis[ans]+1;q.push(noww);} 50 if(noww==b) 51 {printf("%d\n",vis[noww]-1);break;} 52 } 53 } 54 } 55 } 56 return 0; 57 }?
第五題：
一道判斷是不是連通的，用并查集就可以判斷，但是要做一點小處理，存儲的時候如果就一個數字就不管他，如果有兩個或兩個以上就用第一個數字當父親，讓后面其他的數字與其關聯。
E - The Suspects Time Limit:1000MS???? Memory Limit:20000KB???? 64bit IO Format:%I64d & %I64u Submit Status Practice POJ 1611Description
Severe acute respiratory syndrome (SARS), an atypical pneumonia of unknown aetiology, was recognized as a global threat in mid-March 2003. To minimize transmission to others, the best strategy is to separate the suspects from others.In the Not-Spreading-Your-Sickness University (NSYSU), there are many student groups. Students in the same group intercommunicate with each other frequently, and a student may join several groups. To prevent the possible transmissions of SARS, the NSYSU collects the member lists of all student groups, and makes the following rule in their standard operation procedure (SOP).
Once a member in a group is a suspect, all members in the group are suspects.
However, they find that it is not easy to identify all the suspects when a student is recognized as a suspect. Your job is to write a program which finds all the suspects.
Input
The input file contains several cases. Each test case begins with two integers n and m in a line, where n is the number of students, and m is the number of groups. You may assume that 0 < n <= 30000 and 0 <= m <= 500. Every student is numbered by a unique integer between 0 and n?1, and initially student 0 is recognized as a suspect in all the cases. This line is followed by m member lists of the groups, one line per group. Each line begins with an integer k by itself representing the number of members in the group. Following the number of members, there are k integers representing the students in this group. All the integers in a line are separated by at least one space.A case with n = 0 and m = 0 indicates the end of the input, and need not be processed.
Output
For each case, output the number of suspects in one line.Sample Input
100 4 2 1 2 5 10 13 11 12 14 2 0 1 2 99 2 200 2 1 5 5 1 2 3 4 5 1 0 0 0Sample Output
4 1 1下附代碼： 1 #include<iostream> 2 #include<stdio.h> 3 #include<string.h> 4 #include<cmath> 5 #include<algorithm> 6 #include<queue> 7 #define INF 1000005 8 using namespace std; 9 int f[30005],mark[30005],mark1[30005]={0}; 10 int finding(int x) 11 { 12 if(x!=f[x]) 13 f[x]=finding(f[x]); 14 return f[x]; 15 } 16 int together(int x,int y) 17 { 18 int f1=finding(x); 19 int f2=finding(y); 20 if(f1!=f2) 21 {f[f2]=f1;return 1;} 22 return 0; 23 } 24 int main() 25 { 26 int i,m,n,ans,t,bg,j,num,k,h=0,cot=0,sum,late,temp; 27 while(scanf("%d%d",&n,&m)&&(n!=0||m!=0)) 28 { 29 for(i=0;i<=n;i++) 30 f[i]=i; 31 k=0; 32 while(m--) 33 { 34 scanf("%d",&num); 35 if(num>1) 36 { 37 scanf("%d",&bg); 38 if(!mark1[bg]) 39 {mark[k++]=bg;mark1[bg]=1;} 40 for(i=1;i<num;i++) 41 { 42 scanf("%d",&late); 43 together(bg,late); 44 if(!mark1[late]) 45 {mark[k++]=late;mark1[late]=1;} 46 } 47 } 48 else if(num==1) 49 {scanf("%d",&temp); 50 if(!mark1[temp]) 51 {mark[k++]=temp;mark1[temp]=1;}} 52 } 53 sum=0; 54 for(i=0;i<k;i++) 55 { 56 if(finding(0)==finding(mark[i])) 57 sum++; 58 } 59 if(sum==0) 60 printf("1\n"); 61 else 62 printf("%d\n",sum); 63 for(i=0;i<k;i++) 64 mark1[mark[i]]=0; 65 } 66 return 0; 67 }
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第六題：
一道超級水的最小生成樹，用kruskal就可以。但是注意儲存，只需要上三角就行。之前不小心數組開小了。就超了。。
1 #include<iostream> 2 #include<stdio.h> 3 #include<string.h> 4 #include<cmath> 5 #include<algorithm> 6 #include<queue> 7 #define INF 1000000 8 using namespace std; 9 int f[100005]; 10 struct node 11 { 12 int x; 13 int y; 14 int k; 15 }c[100005]; 16 int cmp(struct node a,struct node b) 17 { 18 return a.k<b.k; 19 } 20 int finding(int x) 21 { 22 if(f[x]!=x) 23 f[x]=finding(f[x]); 24 return f[x]; 25 } 26 int together(int x,int y) 27 { 28 int f1,f2; 29 f1=finding(x); 30 f2=finding(y); 31 if(f1!=f2) 32 {f[f2]=f1;return 1;} 33 return 0; 34 } 35 int main() 36 { 37 int i,m,n,ans,t,a,b,bg,j,num,k,h=0,sum=0; 38 while(scanf("%d",&n)!=EOF) 39 { 40 h=0;ans=0;sum=0; 41 for(i=0;i<100005;i++) 42 f[i]=i; 43 for(i=1;i<=n;i++) 44 for(j=1;j<=n;j++) 45 { 46 scanf("%d",&c[h].k); 47 if(c[h].k!=0) 48 { 49 c[h].x=i;c[h].y=j; 50 h++; 51 } 52 } 53 sort(c,c+h,cmp); 54 for(i=1;i<h;i++) 55 { 56 ans=together(c[i].x,c[i].y); 57 if(ans==1) 58 { 59 sum+=c[i].k; 60 } 61 } 62 printf("%d\n",sum); 63 } 64 return 0; 65 }?
第七題：
這題是最短路徑，用dijkstra會超的原因居然是模板錯了！！。最后只能用floyd反而過了。。無語。。模板害死人啊。
if(s[j]==false&&dis[j]<mining) { u=j;mining=mapp[v][j]; }這句才是真的
G - Tram Time Limit:1000MS???? Memory Limit:30000KB???? 64bit IO Format:%I64d & %I64u Submit Status Practice POJ 1847
Description
Tram network in Zagreb consists of a number of intersections and rails connecting some of them. In every intersection there is a switch pointing to the one of the rails going out of the intersection. When the tram enters the intersection it can leave only in the direction the switch is pointing. If the driver wants to go some other way, he/she has to manually change the switch.When a driver has do drive from intersection A to the intersection B he/she tries to choose the route that will minimize the number of times he/she will have to change the switches manually.
Write a program that will calculate the minimal number of switch changes necessary to travel from intersection A to intersection B.
Input
The first line of the input contains integers N, A and B, separated by a single blank character, 2 <= N <= 100, 1 <= A, B <= N, N is the number of intersections in the network, and intersections are numbered from 1 to N.Each of the following N lines contain a sequence of integers separated by a single blank character. First number in the i-th line, Ki (0 <= Ki <= N-1), represents the number of rails going out of the i-th intersection. Next Ki numbers represents the intersections directly connected to the i-th intersection.Switch in the i-th intersection is initially pointing in the direction of the first intersection listed.
Output
The first and only line of the output should contain the target minimal number. If there is no route from A to B the line should contain the integer "-1".Sample Input
3 2 1 2 2 3 2 3 1 2 1 2Sample Output
0下附代碼： 1 #include<iostream> 2 #include<stdio.h> 3 #include<string.h> 4 #include<cmath> 5 #include<algorithm> 6 #include<queue> 7 #define INF 1000000 8 using namespace std; 9 int mapp[105][105],dis[105]; 10 bool s[105]; 11 int n; 12 void dijkstra(int v) 13 { 14 int i,j; 15 for(i=1;i<=n;i++) 16 { 17 dis[i]=mapp[v][i]; 18 s[i]=false; 19 } 20 s[v]=true; 21 dis[v]=0; 22 for(i=2;i<=n;i++) 23 { 24 int mining=INF,u=v; 25 for(j=1;j<=n;j++) 26 { 27 if(s[j]==false&&dis[j]<mining) 28 { 29 u=j;mining=mapp[v][j]; 30 } 31 } 32 s[u]=true; 33 for(j=1;j<=n;j++) 34 { 35 if(s[j]==false&&mapp[u][j]<INF) 36 if(mapp[u][j]+dis[u]<dis[j]) 37 dis[j]=mapp[u][j]+dis[u]; 38 } 39 } 40 } 41 int main() 42 { 43 int i,m,ans,t,a,b,bg,j,num,k; 44 while(scanf("%d%d%d",&n,&a,&b)!=EOF) 45 {for(i=0;i<105;i++) 46 for(j=0;j<105;j++) 47 mapp[i][j]=INF; 48 for(i=1;i<=n;i++) 49 { 50 scanf("%d",&num); 51 for(j=0;j<num;j++) 52 { 53 scanf("%d",&bg); 54 if(j==0) 55 mapp[i][bg]=0; 56 else 57 mapp[i][bg]=1; 58 } 59 } 60 /*for(i=1;i<=n;i++) 61 {for(j=1;j<=n;j++) 62 printf("%d ",mapp[i][j]); 63 cout<<endl;}*/ 64 dijkstra(a); 65 /*for(j=1;j<=n;j++) 66 for(i=1;i<=n;i++) 67 for(k=1;k<=n;k++) 68 if(mapp[i][j]+mapp[j][k]<mapp[i][k]) 69 mapp[i][k]=mapp[i][j]+mapp[j][k];*/ 70 /*for(i=1;i<=n;i++) 71 printf("%d ",dis[i]); 72 cout<<endl;*/ 73 if(dis[b]==INF) 74 printf("-1\n"); 75 else 76 printf("%d\n",dis[b]); 77 /*if(mapp[a][b]==INF) 78 printf("-1\n"); 79 else 80 printf("%d\n",mapp[a][b]);*/ 81 } 82 return 0; 83 }
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第八題：
這題以前做過，就是簡單的素數篩一下，從小到大選擇就好，注意。沒有偶數。。我還調了半天。
J - Goldbach's Conjecture Time Limit:1000MS???? Memory Limit:65536KB???? 64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2262Description
In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture:Every even number greater than 4 can be
written as the sum of two odd prime numbers.
For example:
8 = 3 + 5. Both 3 and 5 are odd prime numbers.
20 = 3 + 17 = 7 + 13.
42 = 5 + 37 = 11 + 31 = 13 + 29 = 19 + 23.
Today it is still unproven whether the conjecture is right. (Oh wait, I have the proof of course, but it is too long to write it on the margin of this page.)
Anyway, your task is now to verify Goldbach's conjecture for all even numbers less than a million.
Input
The input will contain one or more test cases.Each test case consists of one even integer n with 6 <= n < 1000000.
Input will be terminated by a value of 0 for n.
Output
For each test case, print one line of the form n = a + b, where a and b are odd primes. Numbers and operators should be separated by exactly one blank like in the sample output below. If there is more than one pair of odd primes adding up to n, choose the pair where the difference b - a is maximized. If there is no such pair, print a line saying "Goldbach's conjecture is wrong."Sample Input
8 20 42 0Sample Output
8 = 3 + 5 20 = 3 + 17 42 = 5 + 37下附代碼： 1 #include<iostream> 2 #include<stdio.h> 3 #include<string.h> 4 #include<cmath> 5 #include<algorithm> 6 #include<queue> 7 #define INF 1000005 8 using namespace std; 9 queue<int>q; 10 int s[1000005]={0},primee[1000005]; 11 void prime() 12 { 13 int i,k=1,cot; 14 for(i=2;i<INF;i++) 15 { 16 if(s[i]==0) 17 primee[k++]=i; 18 cot=2; 19 while(cot*i<INF) 20 { 21 s[i*cot]=1; 22 cot++; 23 } 24 } 25 } 26 int main() 27 { 28 29 int i=1,n; 30 prime(); 31 //for(i=1;i<100;i++) 32 //cout<<primee[i]<<" "; 33 while(scanf("%d",&n)&&n!=0) 34 { 35 i=1; 36 while(s[n-primee[i]]) 37 { 38 i++; 39 } 40 printf("%d = %d + %d\n",n,primee[i],n-primee[i]); 41 } 42 return 0; 43 }
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第九題：
簡單的進位。字符串轉成數字進行加減就好。
K - Primary Arithmetic Time Limit:1000MS???? Memory Limit:65536KB???? 64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2562Description
Children are taught to add multi-digit numbers from right-to-left one digit at a time. Many find the "carry" operation - in which a 1 is carried from one digit position to be added to the next - to be a significant challenge. Your job is to count the number of carry operations for each of a set of addition problems so that educators may assess their difficulty.Input
Each line of input contains two unsigned integers less than 10 digits. The last line of input contains 0 0.Output
For each line of input except the last you should compute and print the number of carry operations that would result from adding the two numbers, in the format shown below.Sample Input
123 456 555 555 123 594 0 0Sample Output
No carry operation. 3 carry operations. 1 carry operation.下附代碼： 1 #include<iostream> 2 #include<stdio.h> 3 #include<string.h> 4 #include<cmath> 5 char a[200005],b[200005]; 6 int main() 7 { 8 int carry,ans,t,s; 9 while(scanf("%s%s",a,b)&&(a[0]!='0'||b[0]!='0')) 10 { 11 carry=0; 12 t=strlen(a)-1;s=strlen(b)-1;ans=0; 13 while(t>=0&&s>=0) 14 { 15 if((a[t]-48)+(b[s]-48)+carry>9) 16 { 17 carry=(a[t]-48)+(b[s]-48)+carry; 18 ans++; 19 carry/=10; 20 } 21 t--;s--; 22 } 23 if(t<0) 24 { 25 while(s>=0) 26 { 27 if((b[s]-48)+carry>9) 28 { 29 carry=(b[s]-48)+carry; 30 ans++;carry/=10;s--; 31 } 32 else 33 break; 34 } 35 } 36 if(s<0) 37 { 38 while(t>=0) 39 { 40 if((a[t]-48)+carry>9) 41 { 42 carry=(a[t]-48)+carry; 43 ans++;carry/=10;t--; 44 } 45 else 46 break; 47 } 48 } 49 if(ans==0) 50 printf("No carry operation.\n"); 51 else if(ans==1) 52 printf("1 carry operation.\n"); 53 else 54 printf("%d carry operations.\n",ans); 55 } 56 return 0; 57 }
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第十題：就是在n的范圍內，互質的分數的個數。其實就是歐拉函數在1-n范圍內求和，打個線性歐拉函數篩就搞定了。 L - Farey Sequence Time Limit:1000MS???? Memory Limit:65536KB???? 64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2478
Description
The Farey Sequence Fn for any integer n with n >= 2 is the set of irreducible rational numbers a/b with 0 < a < b <= n and gcd(a,b) = 1 arranged in increasing order. The first few areF2 = {1/2}
F3 = {1/3, 1/2, 2/3}
F4 = {1/4, 1/3, 1/2, 2/3, 3/4}
F5 = {1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5}
You task is to calculate the number of terms in the Farey sequence Fn.
Input
There are several test cases. Each test case has only one line, which contains a positive integer n (2 <= n <= 10 6). There are no blank lines between cases. A line with a single 0 terminates the input.Output
For each test case, you should output one line, which contains N(n) ---- the number of terms in the Farey sequence Fn.Sample Input
2 3 4 5 0Sample Output
1 3 5 9下附代碼： 1 #include<iostream> 2 #include<stdio.h> 3 #include<string.h> 4 #include<cmath> 5 #include<algorithm> 6 #include<queue> 7 #define INF 1000005 8 using namespace std; 9 int target[INF],phi[INF]; 10 bool pri[INF]; 11 int main() 12 { 13 int i,m,n,ans,t,bg,j,num,k,h=0,cot=0; 14 long long int sum=0; 15 for(i=2;i<INF;i++) 16 { 17 if(pri[i]==false) 18 { 19 target[cot++]=i; 20 phi[i]=i-1; 21 } 22 for(j=0;j<cot&&i*target[j]<INF;j++) 23 { 24 pri[i*target[j]]=true; 25 if(i%target[j]==0) 26 phi[i*target[j]]=phi[i]*target[j]; 27 else 28 phi[i*target[j]]=phi[i]*(target[j]-1); 29 } 30 } 31 while(scanf("%d",&n)&&n!=0) 32 { 33 sum=0; 34 for(i=2;i<=n;i++) 35 sum+=phi[i]; 36 printf("%lld\n",sum); 37 } 38 return 0; 39 }
其實還有兩題。但都是比較困難的題目了。。待以后有余力的話繼續補充。
又結束了一次訓練，感覺還是沒什么長進，希望暑假能提高代碼能力。。加油！
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?轉載于:https://www.cnblogs.com/linminxuan/p/4559642.html
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