Topic links: https://ac.nowcoder.com/acm/contest/1107/D
The meaning of problems: there is a ring made of balls n + m, n Bobo desirable dyed black balls, the balls m colored white. Bobo right grouping of adjacent balls of the same color, he will re-colored length determined as the product group. He might want to know the color of the sum of weights. The answer to 1e9 + 7 mod.
Problem-solving ideas: personally feel that this question difficult. First, we find the same number painted in black and white areas. Then we can define dp [i] [j] of the j-th ball is divided into blocks of all possible i and the product. Then the state transition equation is:
dp [i] [j] = (ji + 1) * dp [i-1] [i-1] + (ji) * dp [i] [i-1] +. . . + 2 * dp [i-1] [j-2] + dp [i-1] [j-1]. Then we can consider each case the position of the minimum may be painted black piece found a common n + m possible, the final answer can be obtained. Be careful not to arbitrarily modulo, or may time out QAQ. . .
#include<bits/stdc++.h>
using namespace std;
const int maxn=5e3+5;
int dp[maxn][maxn];
int inv[maxn];
const long long mod=1e9+7;
int main(){
inv[1]=1;
for(int i=2;i<=5000;i++){
inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
}
dp[0][0]=1;
for(int i=1;i<=5000;i++){
int tem1=dp[i-1][i-1];
int tem2=1ll*dp[i-1][i-1]*(i-1)%mod;
for(int j=i;j<=5000;j++){
dp[i][j]=(1ll*tem1*j-tem2+mod)%mod;
tem1=(tem1+dp[i-1][j])%mod;
tem2=(tem2+1ll*j*dp[i-1][j])%mod;
}
}
int n,m;
while(scanf("%d%d",&n,&m)!=EOF){
int tem=min(n,m);
long long ans=0;
for(int i=1;i<=tem;i++){
ans=(ans+1ll*dp[i][n]*dp[i][m]%mod*inv[i])%mod;
}
printf("%lld\n",1ll*ans*(n+m)%mod);
}
return 0;
}