思路:
dp[i][j] 表示到第 i 个球为止放了 j 个蓝球的方案数
第 i 个球来自的位置的最右边是min(i, n)
转移方程看代码
代码:
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #include<bits/stdc++.h> using namespace std; #define y1 y11 #define fi first #define se second #define pi acos(-1.0) #define LL long long //#define mp make_pair #define pb push_back #define ls rt<<1, l, m #define rs rt<<1|1, m+1, r #define ULL unsigned LL #define pll pair<LL, LL> #define pli pair<LL, int> #define pii pair<int, int> #define piii pair<pii, int> #define pdd pair<double, double> #define mem(a, b) memset(a, b, sizeof(a)) #define debug(x) cerr << #x << " = " << x << "\n"; #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); //head const int N = 2e3 + 10; const int MOD = 998244353; char s[N]; int dp[N*2][N*2]; int sum[N], sm[N]; int main() { int n; scanf("%s", s+1); n = strlen(s+1); for (int i = 1; i <= n; ++i) sum[i] = sum[i-1] + s[i]-‘0‘; dp[0][0] = 1; for (int i = 1; i <= 2*n; ++i) { int p = min(i, n); for (int j = 0; j < i; j++) { if(sum[p] > j) dp[i][j+1] = (dp[i][j+1] + dp[i-1][j]) % MOD; if(2*i - sum[p] > i-1-j) dp[i][j] = (dp[i][j] + dp[i-1][j]) % MOD; } } printf("%d\n", dp[2*n][sum[n]]); return 0; }
原文地址:https://www.cnblogs.com/widsom/p/10363216.html
时间: 2024-10-30 04:48:38