bzoj1531

背包+倍增

直接背包跑不过去，那么我们把容量分成二进制，然后原来需要枚举c次就只用枚举log(c)次了，这样还是能组合出任意小于等于c的组合方案

#include<bits/stdc++.h>
using namespace std;
const int N = ;
int n, s;
int b[N], c[N], dp[N];
int main()
{
    scanf("%d", &n);
    for(int i = ; i <= n; ++i) scanf("%d", &b[i]);
    for(int i = ; i <= n; ++i) scanf("%d", &c[i]);
    scanf("%d", &s);
    memset(dp, 0x3f3f, sizeof(dp));
    dp[] = ;
    for(int i = ; i <= n; ++i)
    {
        for(int j = ; j <= c[i]; j <<= )
        {
            for(int k = s; k >= j * b[i]; --k) dp[k] = min(dp[k], dp[k - j * b[i]] + j);
            c[i] -= j;
        }
        if(c[i]) for(int j = s; j >= c[i] * b[i]; --j) dp[j] = min(dp[j], dp[j - c[i] * b[i]] + c[i]);
    }
    printf("%d\n", dp[s]);
    return ;
}