Bzoj1042: [HAOI2008]硬币购物

题面

传送门

Sol

容斥原理+背包
处理出所有金币无限制条件凑成\(j\)元的方案数
考虑计算
\(c\)只有\(4\)种,可以容斥一波
就是无限制的总方案-\(1\)个硬币超出限制的方案+\(2\)个的-\(3\)个的+\(4\)个的

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1005);

IL int Input(){
    RG int x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

typedef int Arr[_];
int c[4], tot, mx, pw[4] = {1, 2, 4, 8}, s[_], d[4][_];
ll f[_ * 100] = {1}, ans;

int main(RG int argc, RG char *argv[]){
    for(RG int i = 0; i < 4; ++i) c[i] = Input();
    tot = Input();
    for(RG int i = 1; i <= tot; ++i){
        for(RG int j = 0; j < 4; ++j) d[j][i] = (Input() + 1) * c[j];
        s[i] = Input(), mx = max(mx, s[i]);
    }
    for(RG int i = 0; i < 4; ++i)
        for(RG int j = c[i]; j <= mx; ++j) f[j] += f[j - c[i]];
    for(RG int i = 1; i <= tot; ++i){
        ans = 0;
        for(RG int j = 0; j < 16; ++j){
            RG int op = 1, sum = 0;
            for(RG int k = 0; k < 4; ++k)
                if(j & pw[k]) op = -op, sum += d[k][i];
            ans += (sum > s[i]) ? 0 : (1LL * op * f[s[i] - sum]);
        }
        printf("%lld\n", ans);
    }
    return 0;
}
posted @ 2018-03-27 20:42  Cyhlnj  阅读(...)  评论(... 编辑 收藏