caioj 1086 动态规划入门（非常规DP10：进攻策略）

一开始看到题目感觉很难
然后看到题解感觉这题贼简单，我好像想复杂了
就算出每一行最少的资源（完全背包+二分）然后就枚举就好了。
 

#include<cstdio>
#include<algorithm>
#include<cstring>
#define REP(i, a, b) for(int i = (a); i < (b); i++)
using namespace std;

const int MAXN = 212345;
const int MAXM = 112;
int n, m, k;
int w[MAXM], p[MAXM], q[MAXN];
int f[MAXN], list[MAXN];

int main()
{
	scanf("%d%d%d", &m, &n, &k);
	REP(i, 0, m) scanf("%d", &w[i]);
	REP(i, 0, m) scanf("%d", &p[i]);
	REP(i, 0, n) scanf("%d", &q[i]);
	
	REP(i, 0, m) 
		REP(j, w[i], k + 1)
			f[j] = max(f[j], f[j-w[i]]+p[i]);
		
	int head = 0, sum = 0, ans = 0;
	REP(i, 0, n)
	{
		int t = lower_bound(f, f + k + 1, q[i]) - f;//不是二分答案都用stl二分，简单又不会错 
		sum += t; list[i] = t;
		while(sum > k) sum -= list[head++]; //注意这里是while
		ans = max(ans, i - head + 1);
	}
	printf("%d
", ans);
	
	return 0;
}