Codeforces 1027E Inverse Coloring 【DP】

 

Codeforces 1027E Inverse Coloring

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题目链接

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 1 #include
 2 using namespace std;
 3 #define N 1010
 4 #define LL long long
 5 #define Mod 998244353
 6 int n,k;
 7 LL dp[N][N],ans=0;
 8 LL sum[N][N];
 9 int main(){
10     cin>>n>>k;
11     dp[0][0]=1;
12     for(int i=1;i<=n;i++)
13         for(int j=1;j<=i;j++)
14             for(int p=1;p<=j;p++)
15                 dp[i][j]=(dp[i][j]+dp[i-p][min(j,i-p)])%Mod;
16     for(int i=n;i>=1;i--)dp[n][i]=(dp[n][i]-dp[n][i-1]+Mod)%Mod;
17     for(int i=1;i<=n;i++)
18         for(int j=1;j*i)
19             ans=(ans+dp[n][i]*dp[n][j]%Mod)%Mod;
20     ans=(ans*2)%Mod;
21     printf("%lld",ans);
22     return 0;
23 }

 

转载于:https://www.cnblogs.com/dream-maker-yk/p/9676280.html