1.CodeForces 92A

給一堆圍成圈的小朋友發餅干，小朋友為1~n號，第幾號小朋友每次拿多少塊餅干，問最后剩多少餅干

#include <iostream> #include <cstdio> using namespace std;int main() {// freopen("a.in" , "r" , stdin);int n , m;while(scanf("%d%d" , &n , &m) == 2){int tmp = (1 + n ) * n / 2;m %= tmp;for(int i=1 ; i<=n ; i++){if(m >= i) m-=i;else break;}printf("%d\n" , m);}return 0; } View Code

2.CodeForces 96A

用0，1分別代表己方和對方球員，如果有7個及以上的同隊隊員站一起會危險，問是否處于危險狀態

#include <iostream> #include <cstdio> #include <cstring> using namespace std; const int N = 105; char s[N]; int pos[N];int main() {// freopen("a.in" , "r" , stdin);while(scanf("%s" , s) != EOF){int len = strlen(s);for(int i=0 ; i<len ; i++){if(s[i] == '0') pos[i] = 0;else pos[i] = 1;}int flag = pos[0] , cnt = 1;bool ok = true;for(int i=1 ; i<len ; i++){if(pos[i] == flag){cnt++;if(cnt >= 7){ok = false;break;}}else{flag = pos[i];cnt = 1;}}if(ok) puts("NO");else puts("YES");}return 0; } View Code

3.CodeForces 71A

將字符串按某種規則縮寫

#include <iostream> #include <cstdio> #include <cstring> using namespace std; const int N = 105; char s[N];int main() {// freopen("a.in" , "r" , stdin);int T;scanf("%d" , &T);while(T--){scanf("%s" , s);int len = strlen(s);if(len <= 10){printf("%s\n" , s);continue;}printf("%c%d%c\n" , s[0] , len-2, s[len-1]);}return 0; } View Code

4.CodeForces 71B

枚舉所有可能的值就可以了

#include <iostream> #include <cstdio> #include <cstring> using namespace std; const int N = 105; int val[N];int main() {// freopen("a.in" , "r" , stdin);int n , k , t;while(scanf("%d%d%d" , &n , &k , &t) == 3){int tmp = k*n*t , p , q;memset(val , 0 , sizeof(val));for(p=0 ; p<=n ; p++){int flag = 0;for(q=0 ; q<=k ; q++){int tmp1 = 100*p*k + 100 * q;// cout<<"tmp: "<<tmp<<" "<<tmp1<<endl;if(tmp >= tmp1 && tmp<tmp1+100){flag= 1;break;}}if(flag) break;}// cout<<p<<" "<<q<<endl;for(int i=1 ; i<=p ; i++)val[i] = k;val[p+1] = q;for(int i=1 ; i<=n ; i++)if(i == 1) printf("%d" , val[i]);else printf(" %d" ,val[i]);puts("");}return 0; } View Code

5.CodeForces 71C

枚舉所有的因子，然后令每個數都對這個因子取模，觀察模相同的數是否等于總數除以因子

#include <iostream> #include <cstdio> #include <cstring> #include <cmath> using namespace std; const int N = 100005; int vis[N] , mul[N] , cnt , a[N] , tot[N];void cal(int n) {int t = (int)sqrt(n);cnt = 0;if(n % 4==0){mul[cnt++] = 4;}while(n % 2 == 0) n>>=1;for(int i=3 ; i<=t ; i++){if(n % i == 0){mul[cnt++] = i;while(n % i == 0)n/=i;}if(i > n) break;}if(n>1) mul[cnt++] = n; }int main() {// freopen("a.in" , "r" , stdin);int n , x;while(~scanf("%d" , &n)){int k=0;for(int i=0 ; i< n;i++){scanf("%d" , &x);if(x == 1) a[k++] = i;}cal(n);int flag = 0;for(int i=0 ; i<cnt ; i++){int tmp = n/mul[i];memset(tot , 0 , sizeof(tot));for(int j=0 ; j<k ; j++){int pp = a[j] % tmp;tot[pp] ++;if(tot[pp] == mul[i]){flag = 1;break;}}if(flag) break;}if(flag) puts("YES");else puts("NO");}return 0; } View Code

6.CodeForces 71D

純模擬，代碼量略長

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int N = 60;char str[4]; int num[N][N] , vis[N]; int cnt; //記錄可行矩陣的個數struct Node{int x,y; }node[60];Node pos1 , pos2; Node squre1 , squre2;int change1(char c) {if(c == 'C') return 0;else if(c == 'D') return 1;else if(c == 'H') return 2;else if(c == 'S') return 3; }int change2(char c) {if(c == 'A') return 0;else if(c == 'T') return 9;else if(c == 'J') return 10;else if(c == 'Q') return 11;else if(c == 'K') return 12;else {return (int)(c-'1');} }bool ok(int i , int j) {int flag1 = 1 , flag2 = 1;int mod[20];memset(mod, 0 , sizeof(mod));for(int p=0 ; p<3 ; p++){for(int q=0 ; q<3 ; q++){int t = num[i+p][j+q] % 13;if(mod[t]){flag1 = 0;break;}mod[t] = 1;}if(!flag1) break;}if(flag1) return true;int col = num[i][j] / 13;for(int p=0 ; p<3 ; p++){for(int q=0 ; q<3 ; q++){int t = num[i+p][j+q] / 13;if(t != col){flag2 = 0;break;}}if(!flag2) break;}if(flag2) return true;else return false; }bool ok1(Node m1 , Node m2) {if(m1.x > m2.x+2) return true;if(m1.x+2 < m2.x) return true;if(m1.y+2 < m2.y) return true;if(m1.y > m2.y+2) return true;return false; }void get_martrix(int n , int m) {cnt = 0;for(int i=1 ; i<=n-2 ; i++){for(int j=1 ; j<=m-2 ; j++){if(ok(i,j)){node[cnt].x = i;node[cnt++].y = j;}}} }char* intTostring(int x) {char *s = new char [3];int t1 = x/13;if(t1 == 0) s[1] = 'C';else if(t1 == 1) s[1] = 'D';else if(t1 == 2) s[1] = 'H';else if(t1 == 3) s[1] = 'S';int t2 = x%13;if(t2 == 0) s[0] = 'A';else if(t2 == 9) s[0] = 'T';else if(t2 == 10) s[0] = 'J';else if(t2 == 11) s[0] = 'Q';else if(t2 == 12) s[0] = 'K';else {s[0] = (char)(t2+'1');}s[2] = '\0';return s; }int main() {// freopen("a.in" , "r" , stdin);int n , m;while(scanf("%d%d" , &n , &m) == 2){bool flag1 = false , flag2 = false;int ans1 , ans2 , card1 , card2;//記錄最后選中的在第幾號節點pos1.x = pos1.y = pos2.x = pos2.y = 0;memset(vis , 0 , sizeof(vis));for(int i=1 ; i<=n ; i++)for(int j = 1 ; j<=m ; j++){scanf("%s" , str);if(str[1] == '1'){pos1.x = i;pos1.y = j;flag1 = true;}else if(str[1] == '2'){pos2.x = i;pos2.y = j;flag2 = true;}else{num[i][j] = 13*change1(str[1]) + change2(str[0]);vis[num[i][j]] = 1;}}/* for(int i=1 ; i<=n ; i++){for(int j = 1 ; j<=m ; j++)cout<<num[i][j]<<" ";puts("");}*/bool find = false;int time=0;if(flag1) time++;if(flag2) time++;for(int p=0 ; p<52 ; p++){for(int q=0 ; q<52 ; q++){if(p == q) continue;int change = time;if(flag1 && !vis[p]) num[pos1.x][pos1.y] = p,change--;if(flag2 && !vis[q]) num[pos2.x][pos2.y] = q,change--;if(!change){get_martrix(n , m);for(int i=0 ; i<cnt ; i++){for(int j=i+1 ; j<cnt ; j++){if(ok1(node[i] , node[j])){find=true;ans1 = i , ans2 = j;card1 = p , card2 = q;break;}}if(find) break;}}if(find) break;}if(find) break;}if(!find)puts("No solution.");else{puts("Solution exists.");if(!flag1 && !flag2){puts("There are no jokers.");}else if(flag1 && flag2){char *s1 = new char [3];char *s2 = new char [3];s1 = intTostring(card1);s2 = intTostring(card2);printf("Replace J1 with %s and J2 with %s.\n" , s1 , s2);}else if(flag1){char *s1 = new char [3];s1 = intTostring(card1);printf("Replace J1 with %s.\n" , s1);}else if(flag2){char *s1 = new char [3];s1 = intTostring(card2);printf("Replace J2 with %s.\n" , s1);}printf("Put the first square to (%d, %d).\n" , node[ans1].x , node[ans1].y);printf("Put the second square to (%d, %d).\n" , node[ans2].x , node[ans2].y);}}return 0; } View Code

7.CodeForces 88E

一道簡單的nim博弈，sg函數值的計算過程中注意保存得到的最小堆數，過程可以用異或的前綴和幫助自己提高效率

#include <cstdio> #include <cstring> #include <iostream> #include <cmath> using namespace std; const int N = 100010;int sg[N] , sum[N] , minn[N] , vis[N];void init_sg() {sg[0] = sg[1] = sg[2] = 0;sum[0] = sum[1] = sum[2] = 0;memset(minn , 0x3f , sizeof(minn));for(int n=3 ; n<=100000 ; n++){int t = 2*n;int factor = (int)(sqrt(t+0.5));for(int k=2 ; k<=factor ; k++){if(t % k == 0 && t/k+1-k>0 && !((t/k+1-k)&1)){int a = (t/k+1-k)/2;int p = sum[a+k-1]^sum[a-1];vis[p] = n;if(p == 0) minn[n] = min(minn[n] , k);}}//找沒有出現過的最小的sg函數int v = 0;while(1){if(vis[v] != n){sg[n] = v;break;}v++;}//記錄當前的sum[]前綴sum[n] = sum[n-1]^sg[n];} }int main() {// freopen("a.in" , "r" , stdin); init_sg();int n;while(scanf("%d" , &n) == 1){if(!sg[n]){puts("-1");continue;}printf("%d\n" , minn[n]);}return 0; } View Code

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轉載于:https://www.cnblogs.com/CSU3901130321/p/4253354.html

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