POJ 1276 Cash Machine（DP多重背包）

题意：给出一个价值sum，然后给出n，代表n个方案，接着n对代表个数与价值，要求最接近sum，但不超过sum的价值。

思路：多重背包，拆分成二进制转化01背包。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<ctime>
#define eps 1e-6
#define LL long long
#define pii pair<int, int>
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

const int MAXN = 100100;
//const int INF = 0x3f3f3f3f;
int n, v, sz;
bool dp[MAXN];
int a[200];
int main() {
//freopen("input.txt", "r", stdin);
while(scanf("%d", &n) == 1) {
memset(dp, 0, sizeof(dp));
sz = 0;
dp[0] = 1;
int cat; cin >> cat;
for(int i = 1; i <= cat; i++) {
int u, v; scanf("%d%d", &u, &v);
if(!u || !v) continue;
int ind = 0;
while((1<<(ind+1))-1 < u) {
a[++sz] = (1<<ind) * v;
ind++;
}
a[++sz] = (u-(1<<ind)+1) * v;
}
for(int i = 1; i <= sz; i++) {
for(int j = n; j >= a[i]; j--) {
if(dp[j-a[i]]) dp[j] = 1;
}
}
for(int i = n; i >= 0; i--) {
if(dp[i]) {
cout << i << endl;
break;
}
}
}
return 0;
}