Codeforces 1172C2 Nauuo and Pictures (hard version) dp

Nauuo and Pictures (hard version

首先考虑简单版本的， 一个一个dp求出来， 分成三坨， 一坨当前要求照片， 一坨除了当前的喜欢的照片， 一坨除了当前的讨厌的照片。

单次dp   50 ^ 4

感觉hard的也挺简单的。。 

我们先算出最后喜欢的照片的总w， 和讨厌的照片的总w， 然后每个的贡献就是在原先的w中所占的比例。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 3000 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-10;
const double PI = acos(-1);

template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;}

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int power(int a, int b) {
    int ans = 1;
    while(b) {
        if(b & 1) ans = 1LL * ans * a % mod;
        a = 1LL * a * a % mod; b >>= 1;
    }
    return ans;
}

int n, m, a[200007], w[200007], r[200007];
int sumL, sumH, EL, EH;

int dp[N][N];

int main() {

    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &w[i]);
        if(a[i]) sumL += w[i];
        else sumH += w[i];
    }

    EL = sumL;
    EH = sumH;

    dp[0][0] = 1;

    for(int i = 0; i < m; i++) {
        for(int a = 0; a <= i; a++) {
            int b = i - a;
            add(dp[i + 1][a + 1], 1LL * dp[i][a] * (sumL + a) % mod * power(sumL + sumH + a - b, mod - 2) % mod);
            add(dp[i + 1][a], 1LL * dp[i][a] * (sumH - b) % mod * power(sumL + sumH + a - b, mod - 2) % mod);
        }
    }

    for(int a = 0; a <= m; a++) {
        int b = m - a;
        add(EL, 1LL * a * dp[m][a] % mod);
        sub(EH, 1LL * b * dp[m][a] % mod);
    }

    for(int i = 1; i <= n; i++) {
        if(a[i]) {

            r[i] = 1LL * EL * w[i] % mod * power(sumL, mod - 2) % mod;

        } else {

            r[i] = 1LL * EH * w[i] % mod * power(sumH, mod - 2) % mod;

        }
    }

    for(int i = 1; i <= n; i++) printf("%d\n", r[i]);
    return 0;
}

/*
*/