# BZOJ 1042 [HAOI2008]硬币购物(完全背包+容斥)

4种硬币买价值为V的商品，每种硬币有numi个，问有多少种买法

1000次询问，numi<1e5

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>

#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x))

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-6;
const int mod = 1e9+7;
const int maxn = 2e6+100;
const int maxm = 2e6+100;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);

ll dp[maxn];
ll c[5];
ll v[maxn];
ll num[5];
ll ans,V;
//dfs搜容斥组合
void dfs(int x, int k, ll sum){//搜到第x个，已经选了k个，当前组合一共需要减sum
//printf("%d %d %lld\n",x,k,sum);
if(V-sum < 0)return;
if(x==5){
//容斥判断该加还是减
if(k==0)return;
if(k&1) ans += dp[V-sum];
else ans -= dp[V-sum];
return;
}
dfs(x+1, k, sum);//当前不选
dfs(x+1,k+1,sum+c[x]*(num[x]+1));//选
}
int main(){
for(int i = 1; i <= 4; i++){
scanf("%lld", &c[i]);
}
int T;
scanf("%d", &T);
dp[0] = 1;
for(int i = 1; i <= 4; i++){
for(int j = 0; j <= maxn; j++){
if(j-c[i]>=0)dp[j] += dp[j-c[i]];
}
}
while(T--){
for(int i = 1; i <= 4; i++){
scanf("%lld", &num[i]);
}
scanf("%lld", &V);
ans = 0;
dfs(1, 0, 0);
printf("%lld\n",dp[V]-ans);
}
return 0;
}

/*
1 2 5 10 1
3 2 3 1 10

*/

## [HAOI2008]硬币购物否

posted @ 2018-12-13 23:45  wrjlinkkkkkk  阅读(...)  评论(...编辑  收藏