|洛谷|NOIP2007|动态规划|P1095 守望者的逃离
http://www.luogu.org/problem/show?pid=1095
第一次循环让守望者只能闪烁,然后第二次循环让守望者只能跑步,比较两次的长短即可
#include<cstdio> #include<algorithm> #include<cstring> #include<queue> #define ms(i,j) memset(i,j, sizeof i); using namespace std; int f[300005]; int main() { int m,s,t; scanf("%d%d%d", &m, &s, &t); f[0] = 0; for (int i=1;i<=t;i++) { if (m<10) { f[i] = f[i-1]; m += 4; }else {m -= 10; f[i] = f[i-1]+60;} } for (int i=1;i<=t;i++) { f[i] = max(f[i], f[i-1]+17); if (f[i]>=s) { printf("Yes\n%d", i); return 0; } } printf("No\n%d", f[t]); return 0; }