标签:with pictures count() 除了 lse for c++ count nbsp
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; } /* */
Codeforces 1172C2 Nauuo and Pictures (hard version) dp
标签:with pictures count() 除了 lse for c++ count nbsp
原文地址:https://www.cnblogs.com/CJLHY/p/11116659.html