P1782 旅行商的背包

二进制处理多重背包,,,
其实无非就是将物品拆成几组, 每一组的数量都是 (2^n) 个, 然后因为二进制可以组成所有情况所以就不会漏解啦
比余数分组的做法不知道简单到哪里去了

#include <cstdio>
#include <cstring>
#include <cassert>
#include <iostream>
#include <algorithm>
using namespace std;
const int MAXN = 1e4 + 10;
const int MAXW = 1e3 + 10;
inline int read(){
    char ch = getchar(); int x = 0; bool flag = false;
    while(!isdigit(ch)) flag |= (ch == '-'), ch = getchar();
    while(isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return flag ? -x : x;
}

int N, M, C;
int f[MAXN], w[MAXN * 100], v[MAXN * 100];

int main(){
    cin>>N>>M>>C;
    int n = 0;
    for(int i = 1; i <= N; i++) {
        int wei = read(), cost = read(), num = read();
        for(int j = 1; j <= num; j <<= 1) 
            w[++n] = j * wei, v[n] = j * cost, num -= j;
        if(num) w[++n] = wei * num, v[n] = cost * num;
    }

    for(int i = 1; i <= n; i++) 
        for(int j = C; j >= w[i]; --j) 
            f[j] = max(f[j], f[j - w[i]] + v[i]);

    for(int i = 1; i <= M; i++) {
        int a = read(), b = read(), c = read();
        for(int j = C; j; j--)
            for(int k = 0; k <= j; k++)
                f[j] = max(f[j], f[j - k] + a * k * k + b * k + c);
    }
    printf("%d
", f[C]);
    return 0;
}