Problem C?六學家的困惑

https://ac.nowcoder.com/acm/contest/625/C

題解：區間DP

1、對給定的兩個字符串按規則以后最大排列（區間DP）

2、反轉字符串1，與字符串2拼接，得到字符串3

3、初始化DP數組僅字符串1和字符串2的銜接的兩個字符

4、區間DP

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,p,l,r,u,v; int ans,cnt,flag,temp,sum; int a[N]; string str1,str2,str3;struct node{}; int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);scanf("%d",&t);int T=0;while(t--){string dp[3][100][100];//scanf("%d",&n);cin>>str1>>str2;int len1=str1.size();int len2=str2.size();for(int i=0;i<len1;i++){dp[0][i][i]=str1[i];}for(int i=0;i<len2;i++){dp[1][i][i]=str2[i];}for(int i=2;i<=len1;i++){for(int j=0;j+i-1<len1;j++){int l=j;int r=j+i-1;//cout<<dp[0][l+1][r]+str1[l]<<endl;if(str1[r]+dp[0][l][r-1]<str1[l]+dp[0][l+1][r]){dp[0][l][r]=str1[l]+dp[0][l+1][r];}else{dp[0][l][r]=str1[r]+dp[0][l][r-1];}//cout<<dp[0][l][r]<<endl;}}for(int i=2;i<=len2;i++){for(int j=0;j+i-1<len2;j++){int l=j;int r=j+i-1;if(str2[r]+dp[1][l][r-1]<str2[l]+dp[1][l+1][r]){dp[1][l][r]=str2[l]+dp[1][l+1][r];}else{dp[1][l][r]=str2[r]+dp[1][l][r-1];}}}for(int i=0;i<len1+len2;i++){dp[2][i][i]="";}reverse(&dp[0][0][len1-1][0],&dp[0][0][len1-1][len1]); // cout<<dp[0][0][len1-1]<<endl; // cout<<dp[1][0][len2-1]<<endl;str3=dp[0][0][len1-1]+dp[1][0][len2-1];dp[2][len1-1][len1-1]=dp[0][0][len1-1][len1-1];dp[2][len1][len1]=dp[1][0][len2-1][0]; // cout<<str3<<endl; // cout<<dp[2][len1-1][len1-1]<<" "<<dp[2][len1][len1]<<endl;for(int i=2;i<=len1+len2;i++){for(int j=0;j+i-1<len1+len2;j++){int l=j;int r=j+i-1;if(dp[2][l+1][r]!=""&&dp[2][l][r-1]!=""){if(dp[2][l][r-1]+str3[r]<dp[2][l+1][r]+str3[l]){dp[2][l][r]=dp[2][l+1][r]+str3[l];}else{dp[2][l][r]=dp[2][l][r-1]+str3[r];}}else if(dp[2][l+1][r]!=""||dp[2][l][r-1]!=""){if(dp[2][l+1][r]!="")dp[2][l][r]=dp[2][l+1][r]+str3[l];elsedp[2][l][r]=dp[2][l][r-1]+str3[r];}//cout<<dp[2][l][r]<<endl;}}printf("Case #%d: ",++T);cout<<dp[2][0][len1+len2-1]<<endl;}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

?

Problem E?數獨挑戰

https://ac.nowcoder.com/acm/contest/625/E

題解：數獨游戲裸題

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int using namespace std; typedef long long ll; typedef __int128 lll; const int N=10000; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,q; ll ans=-1; struct node {int lx , sum ; }line[ 11 ] ;//記錄每行需要填的零的個數; int a[10][10]; int b[10][10]; bool row[10][10]; bool col[10][10]; bool block[10][10]; void dfs(int id,int x,int y){if(id==9&&y==9){for(int i=1;i<=9;i++){for(int j=1;j<=9;j++){cout << a[i][j]<< " ";}cout << endl;}return;}if(y==9){y=1;x=line[++id].lx;}else{y++;}if(!a[x][y]){for(int i=1;i<=9;i++){if(!row[x][i]&&!col[y][i]&&!block[b[x][y]][i]){row[x][i]=1;col[y][i]=1;block[b[x][y]][i]=1;a[x][y]=i;dfs(id,x,y);a[x][y]=0;row[x][i]=0;col[y][i]=0;block[b[x][y]][i]=0;}}}else{dfs(id,x,y);}} int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//scanf("%d",&n);for(int i=1;i<=3;i++){for(int j=1;j<=3;j++){t=(i-1)*3+j;int r=(i-1)*3;int c=(j-1)*3;for(int k=1;k<=3;k++){for(int l=1;l<=3;l++){b[r+k][c+l]=t;}}}}for(int i=1;i<=9;i++){line[i].lx=i;line[i].sum=0;for(int j=1;j<=9;j++){scanf("%d",&a[i][j]);if(a[i][j]){line[i].sum++;row[i][a[i][j]]=1;col[j][a[i][j]]=1;block[b[i][j]][a[i][j]]=1;}}}dfs(1,line[1].lx,0);//cout << "Hello world!" << endl;return 0; }

Problem F?翻牌游戲

https://ac.nowcoder.com/acm/contest/625/F

題解：規律

C++版本一

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,p,l,r,u,v; int ans,cnt,flag,temp,sum; int a[N]; char str; struct node{}; int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);scanf("%d",&t);while(t--){scanf("%d",&n);printf("%.2f\n",(double)2*n-1);}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

C++版本二

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int MAXN = 5005; double arr[MAXN][MAXN]; int t, n; void Init(){arr[1][1] = 1.0;arr[2][2] = 1.0 / 3.0;arr[2][3] = arr[2][2];arr[2][4] = arr[2][3];for(int i = 3; i <= 5000; i ++){for(int j = i; j <= 2 * i; j ++){if(j == 2 * i){double ans = 1.0;for(int z = i; z < 2 * i; z ++){ans -= arr[i][z];}arr[i][j] = ans;}else if(j == i){double ans = (2.0 * i) / (2.0 * i * (2.0 * i - 1.0));///1/5arr[i][j] = arr[i - 1][i - 1] * ans;}else{double ans = (2.0 * i) / (2.0 * i * (2.0 * i - 1.0));arr[i][j] = ans * arr[i - 1][j - 1] * i;}}} } int main() {scanf("%d", &t);Init();while(t --){scanf("%d", &n);double re = 0.0;for(int i = n; i <= 2 * n; i ++){re += (arr[n][i] * i * 1.0);}printf("%.0f.00\n", re);}return 0; }

Problem H?Parco_Love_GCD

https://ac.nowcoder.com/acm/contest/625/H

C++版本一

題解：遞推

#include <bits/stdc++.h> using namespace std; #define LL long long #define FI first #define SE second #define PB push_back typedef pair<LL,LL> PII; const int N=1e6+7,INF=1e9,mod=1e9+7; int n,m; LL a[N]; vector<PII>v[N]; int main() {cin>>n;LL ans=0;for(int i=1;i<=n;i++){scanf("%lld",&a[i]);ans=(ans+a[i])%mod;if(i==1){v[i].PB(make_pair(a[i],1));continue;}v[i].PB(make_pair(a[i],i));/*if(a[i]!=v[i-1][0].FI){v[i].PB(make_pair(a[i],i));}*/int l=v[i-1].size();LL st=i;for(int j=0;j<l;j++){LL k=__gcd(v[i-1][j].FI,a[i]);if(k==1){ans=(ans+st-1)%mod;v[i].PB(make_pair(1,1));break;}ans=(ans+k*(st-v[i-1][j].SE))%mod;st=v[i-1][j].SE;if(k==v[i][v[i].size()-1].FI){v[i][v[i].size()-1].SE=v[i-1][j].SE;}else v[i].PB(make_pair(k,v[i-1][j].SE));}//if(a[i]%v[i-1][0].FI==0)continue;}cout<<ans<<endl;return 0; }

C++版本二

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=500000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,p,l,r,u; ll ans,cnt,flag,temp,sum; int a[N]; char str; struct node{int val,id;node(){};node(int _val,int _id):val(_val),id(_id){} }; vector<node>v[N]; int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%d",&a[i]);v[i].push_back({a[i],i});ans=(ans+a[i])%MOD;if(i==1)continue;int len=v[i-1].size();int pos=i;for(int j=0;j<len;j++){int k=__gcd(v[i-1][j].val,a[i]);if(k==1){ans=(ans+pos-1)%MOD;v[i].push_back({1,1});break;}ans=(ans+(ll)k*(pos-v[i-1][j].id))%MOD;pos=v[i-1][j].id;if(k==v[i][v[i].size()-1].val){v[i][v[i].size()-1].id=v[i-1][j].id;}else v[i].push_back({k,v[i-1][j].id});}}cout<<ans<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

?

Problem I?炒股

https://ac.nowcoder.com/acm/contest/625/I

題解：按低買高賣的原則

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=500000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,p,l,r,u,v; ll ans,cnt,flag,temp,sum; int a[N]; char str; struct node{}; int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%d",&a[i]);}int pos=1;int maxl=a[1];for(int i=2;i<=n;i++){if(a[i]<maxl){ans+=maxl-a[pos];pos=i;}maxl=a[i];}ans+=maxl-a[pos];cout<<ans<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

?

?

Problem L?石頭剪刀布

https://ac.nowcoder.com/acm/contest/625/L

題解：

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,p,l,r,u,v; int ans,cnt,flag,temp,sum; int a[N]; char str[N]; struct node{}; int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);scanf("%d",&t);while(t--){//scanf("%d",&n);cin>>str;if(str[0]=='S'){cout<<"Rock"<<endl;}else if(str[0]=='R'){cout<<"Paper"<<endl;}else{cout<<"Scissors"<<endl;}}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

?

總結

以上是生活随笔為你收集整理的2019年华南理工大学程序设计竞赛（春季赛）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。