TYVJ 1108 守望者的逃离 解题报告

　　DP，DP方程好复杂，不知道怎么说的好，代码里分解吧：

#include <stdio.h>
#include <stdlib.h>
#define max(a, b) ((a)>(b)?(a):(b))
int g[300001][2], f[300001];

int main(int argc, char **argv)
{
	int i, min = -1;
	int m, s, t;
	scanf("%d%d%d", &m, &s, &t);
	g[0][0] = m;
	for(i = 1; i <= t; i++){
		if(g[i - 1][0] >= 10){
			g[i][0] = g[i - 1][0] - 10;
			g[i][1] = g[i - 1][1] + 60;
		}else{
			g[i][0] = g[i - 1][0] + 4;
			g[i][1] = g[i - 1][1];
		}
		f[i] = max(f[i - 1] + 17, g[i][1]);
		if(min == -1 && f[i] >= s){
			min = i;
		}
	}
	if(min == -1){
		printf("No\n%d\n", f[t]);
	}else{
		printf("Yes\n%d\n", min);
	}
	return 0;
}