标签：strlen ext 模拟 不难 amp #define 简单 close lun
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简单做了一下，前三题比较水，第四题应该算是经典题
直接暴力枚举即可
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int MAXN = 1001, INF = 1e9 + 10; inline int read() { char c = getchar(); int x = 0, f = 1; while(c < ‘0‘ || c > ‘9‘) {if(c == ‘-‘) f = -1; c = getchar();} while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = getchar(); return x * f; } int N, ans = 0; char s[MAXN]; main() { #ifdef WIN32 freopen("a.in", "r", stdin); #endif N = read(); for(int i = 1; i <= N; i++) { scanf("%s", s + 1); int L = strlen(s + 1); for(int j = 1; j <= L - 1; j++) { if((s[j] == ‘c‘ && s[j + 1] == ‘h‘)|| (s[j] == ‘h‘ && s[j + 1] == ‘e‘)|| (s[j] == ‘e‘ && s[j + 1] == ‘f‘)) {ans++; break;} } } printf("%d", ans); }
维护一个所有元素的和，直接模拟即可
#include<cstdio> #include<cstring> #include<algorithm> #define int long long using namespace std; const int MAXN = 1e6 + 10, INF = 1e9 + 10; inline int read() { char c = getchar(); int x = 0, f = 1; while(c < ‘0‘ || c > ‘9‘) {if(c == ‘-‘) f = -1; c = getchar();} while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = getchar(); return x * f; } int N, K, ans = 0; int a[MAXN]; main() { #ifdef WIN32 freopen("a.in", "r", stdin); #endif int T = read(); while(T--) { N = read(); K = read(); int sum = 0, ans = 0; for(int i = 1; i <= N; i++) a[i] = read(), sum += a[i]; for(int i = 1; i <= N; i++) if(a[i] + K > sum - a[i]) ans++; printf("%d\n", ans); } }
把式子化成$2B = A +C$的形式，不难看出改B一定是最优的。
特判一下奇偶性即可
#include<cstdio> #include<cstring> #include<algorithm> #define int long long using namespace std; const int MAXN = 1e6 + 10, INF = 1e9 + 10; inline int read() { char c = getchar(); int x = 0, f = 1; while(c < ‘0‘ || c > ‘9‘) {if(c == ‘-‘) f = -1; c = getchar();} while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = getchar(); return x * f; } main() { #ifdef WIN32 freopen("a.in", "r", stdin); #endif int N = read(); while(N--) { int A = read(), B = read(), C = read(); int ans = abs(2 * B - A - C); if(ans & 1) printf("%lld\n", ans / 2 + 1); else printf("%lld\n", ans / 2); } }
个人感觉是一道比较好的题
设所有元素的和为$sum$
不难发现，不论如何分，分成的段数一定是$sum$的因子
而且不论如何分，第一段一定是$1-x$（以$1$为起点）
这样我们遇到一个因子就枚举一边序列暴力分割就可以
这题TM居然卡常
#include<cstdio> #include<cstring> #include<algorithm> #define int long long using namespace std; const int MAXN = 1e6 + 10, INF = 1e9 + 10; inline int read() { char c = getchar(); int x = 0, f = 1; while(c < ‘0‘ || c > ‘9‘) {if(c == ‘-‘) f = -1; c = getchar();} while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = getchar(); return x * f; } int a[MAXN]; char ans[MAXN]; main() { #ifdef WIN32 freopen("a.in", "r", stdin); #endif int T = read(); while(T--) { int N = read(), sum = 0; for(int i = 1; i <= N; i++) a[i] = read(), sum += a[i]; for(int i = 1; i <= N; i++) { if(sum % i != 0) {ans[i] = ‘0‘; continue;} int cur = 0, num = 0; for(int j = 1; j <= N; j++) { cur += a[j]; if(cur == sum / i) cur = 0; else if(cur > sum / i) {ans[i] = ‘0‘; break;} } ans[i] = (cur == 0 ? ‘1‘ : ‘0‘); } for(int i = 1; i <= N; i++) putchar(ans[i]); puts(""); } }
CodeChef March Lunchtime 2018 div2
标签：strlen ext 模拟 不难 amp #define 简单 close lun
原文地址：https://www.cnblogs.com/zwfymqz/p/9183338.html
