hdu 3398

String

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2161    Accepted Submission(s): 628

Problem Description

Recently,
lxhgww received a task : to generate strings contain '0's and '1's
only, in which '0' appears exactly m times, '1' appears exactly n times.
Also, any prefix string of it must satisfy the situation that the
number of 1's can not be smaller than the number of 0's . But he can't
calculate the number of satisfied strings. Can you help him?

 

Input

T(T<=100) in the first line is the case number.
Each case contains two numbers n and m( 1 <= m <= n <= 1000000 ).

 

Output

Output the number of satisfied strings % 20100501.

 

Sample Input

1

2 2

 

Sample Output

2

 

Author

lxhgww

 

Source

HDOJ Monthly Contest – 2010.05.01

 

Recommend

lcy   |   We have carefully selected several similar problems for you:  3400 3402 3401 3399 3404

 

转化成Cn+m n - Cn+m n+1

 #include<iostream>
 #include<stdio.h>
 #include<cstring>
 #include<cstdlib>
 using namespace std;
 typedef long long LL;
 const long long mod = ;

 bool s[];
 int prime[],len;
 void Init(){
     int i,j;
     memset(s,false,sizeof(s));
     len=;
     for(i=;i<=;i++)
     {
         if(s[i]==true) continue;
         prime[++len]=i;
         if(i>) return;
         for(j=i*;j<=;j=j+i)
         s[j]=true;
     }
 }
 int get_num(int n,int m){
     int ans=;
     while(n){
         n=n/m;
         ans=ans+n;
     }
     return ans;
 }
 LL pow_mod(LL a,LL b)
 {
     LL ans=;
     while(b)
     {
         if(b&){
             ans=(ans*a)%mod;
         }
         b=b>>;
         a=(a*a)%mod;
     }
     return ans;
 }

 void solve(int n,int m){
     int ans;
     LL sum1=,sum2=;
     for(int i=;i<=len;i++)
     {
         if(prime[i]>n+m)break;
         ans=get_num(n+m,prime[i])-get_num(n,prime[i])-get_num(m,prime[i]);
         sum1=(sum1*pow_mod(prime[i],ans))%mod;

         ans=get_num(n+m,prime[i])-get_num(n+,prime[i])-get_num(m-,prime[i]);
         sum2=(sum2*pow_mod(prime[i],ans))%mod;
     }
    // printf("%lld %lld\n",sum1,sum2);
     if(sum1<sum2) sum1=sum1-sum2+mod;
     else sum1=sum1-sum2;
     printf("%lld\n",sum1);
 }
 int main()
 {
     Init();
     int T,n,m;
     scanf("%d",&T);
     while(T--){
         scanf("%d%d",&n,&m);
         solve(n,m);
     }
     return ;
 }