省选前集训 状压dp（容斥原理）

Kykneion asma

Time limit :1000ms

On the last day before the famous mathematician Swan's death, he left a problem to the world: Given integers n and ai for 0≤i≤4, calculate the number of n-digit integers which have at most ai-digit i in its decimal representation (and have no 5,6,7,8 or 9). Leading zeros are not allowed in this problem.

直接统计较为困难

考虑容斥：所有情况-至少一个超出+至少两个超出-.....

然后就可以状压dp了。dp[i][j]表示第i为状态为j，从低向高dp，每次可以把已经超出了的和不在意的任意填。或是枚举当前已超出的为在哪超出。乘组合数转移。

注意容斥思想的应用，直接统计较难时，先统计一部分，在减去重了的

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define maxn 20020
#define mod 1000000007

typedef long long LL;
LL dp[maxn][50];
LL fac[maxn],inv[maxn],cnt[maxn];
int ans;
int n,a[5];

inline int power(LL x,int y){
    LL res = 1;
    while ( y ){
        if ( y & 1 ) res = (res * x) % mod;
        x = (x * x) % mod;
        y >>= 1;
    }
    return res % mod;
}
void init(){
    fac[0] = 1;
    for (int i = 1 ; i <= n ; i++){
        fac[i] = (fac[i - 1] * (LL) i) % mod;;
        inv[i] = power(fac[i],mod - 2);
    }
    for (int i = 1 ; i < 32 ; i++){
        int now = i;
        while ( now ) {if ( now & 1 ) cnt[i]++; now >>= 1;}
    //    cout<<cnt[i]<<" ";
    }
    //cout<<endl;
}
inline LL C(int n,int m){
    if ( n == m || m == 0 ) return 1;
    if ( m > n ) return 0;
    return (fac[n] * inv[m] % mod * inv[n - m] % mod) % mod;
}
int doDp(){
    int ans = power(5,n - 1);
    for (int i = 1 ; i <= 5 ; i++){ //枚举状态中必须超出的数量
        memset(dp,0,sizeof(dp));
        dp[0][0] = 1;
        for (int j = 1 ; j <= n - 1 ; j++){
            for (int k = 0 ; k < 32 ; k++){
                if ( cnt[k] > i ) continue;
                dp[j][k] = (dp[j][k] + (dp[j - 1][k] * (LL) (5 - i + cnt[k]))) % mod;
                for (int l = 0 ; l <= 4 ; l++){
                    if ( ((k >> l) & 1) && (j > a[l]) ) 
                        dp[j][k] = (dp[j][k] + dp[j - a[l] - 1][(1 << l) ^ k] * C(j - 1,a[l])) % mod;
                }
            }
        }
        if ( i & 1 ){
            for (int j = 0 ; j < 32 ; j++) if ( i == cnt[j] ) ans = (int) ((LL)ans - dp[n - 1][j]) % mod;
        }
        else{
            for (int j = 0 ; j < 32 ; j++) if ( i == cnt[j] ) ans = (int) ((LL)ans + dp[n - 1][j]) % mod;
        }
    }
    return (ans % mod + mod) % mod; 
}
int main(){
    scanf("%d",&n);
    for (int i = 0 ; i <= 4 ; i++) scanf("%d",&a[i]);
    init();
    for (int i = 1 ; i <= 4 ; i++) if ( a[i] ) a[i]-- , ans = (ans + doDp()) % mod , a[i]++;//cout<<ans<<endl;
    printf("%d
",ans);
    return 0;
}