BZOJ 1042 [HAOI2008]硬币购物(完全背包+容斥)

题意：

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

思路：

完全背包计算出没有numi限制下的买法，

然后答案为dp[V]-（s1+s2+s3+s4）+（s12+s13+s14+s23+s24+s34）-（s123+s124+s134+s234）+s1234

其中s...为某硬币超过限制的方案数

求s的方法：

如s1:硬币1超过限制，就是硬币1至少选了num1+1个，其他随便，所以s1=dp[V-c1*(num1+1)]

同理s12 = dp[V - c1 * (num1 + 1) - c2 * (num2 + 1)]

代码：

#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
",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
",dp[V]-ans);
    }
    return 0;
}

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

 */

[HAOI2008]硬币购物否