标签：printf ide should frame and from nal single paint
InputThere might be multiple test cases, no more than 5. You need to read till the end of input.
For each test case, a line containing four integers n,m,A,B.
1≤n,m,A,B≤3000 1≤n,m,A,B≤3000
.
OutputFor each test case, output a line containing the answer modulo 998244353.
Sample Input
3 4 1 2
Sample Output
169
题意：给N*M的空白格子染色，求至少x行，至少y列被染色的方案数。
思路：不会，占位。
#include<bits/stdc++.h> using namespace std; #define LL long long const int maxn = 3001; const int MOD = 998244353; int n, m, A, B, ans; int C[maxn][maxn], two[maxn * maxn]; int fa[maxn], fb[maxn]; void Init() { for(int i = 0; i < maxn; ++i) { for(int j = 0; j <= i; ++j) { if(j == i || j == 0) { C[i][j] = 1; } else { C[i][j] = C[i - 1][j - 1] + C[i - 1][j]; if(C[i][j] >= MOD) { C[i][j] -= MOD; } } } } two[0] = 1; for(int i = 1; i < maxn * maxn; ++i) { two[i] = two[i - 1] * 2; if(two[i] >= MOD) { two[i] -= MOD; } } } int main() { Init(); while(~scanf("%d%d%d%d", &n, &m, &A, &B)) { ans = 0; for(int i = A; i <= n; ++i) { fa[i] = 0; for(int j = A; j < i; ++j) { fa[i] = (fa[i] + (LL)C[i][j] * fa[j]) % MOD; } fa[i] = 1 - fa[i]; if(fa[i] < 0) { fa[i] += MOD; } } for(int i = B; i <= m; ++i) { fb[i] = 0; for(int j = B; j < i; ++j) { fb[i] = (fb[i] + (LL)C[i][j] * fb[j]) % MOD; } fb[i] = 1 - fb[i]; if(fb[i] < 0) { fb[i] += MOD; } } for(int i = A; i <= n; ++i) { LL tmp = (LL)fa[i] * C[n][i] % MOD; for(int j = B; j <= m; ++j) { ans = (ans + ((tmp * fb[j] % MOD) * C[m][j] % MOD) * two[(n - i) * (m - j)] % MOD) % MOD; } } printf("%d\n", ans); } return 0; }
标签：printf ide should frame and from nal single paint
原文地址：https://www.cnblogs.com/hua-dong/p/9851697.html
