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

 

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转化成Cn+m n - Cn+m n+1

 1 #include<iostream>

 2 #include<stdio.h>

 3 #include<cstring>

 4 #include<cstdlib>

 5 using namespace std;

 6 typedef long long LL;

 7 const long long mod = 20100501;

 8 

 9 bool s[2000010];

10 int prime[150000],len;

11 void Init(){

12     int i,j;

13     memset(s,false,sizeof(s));

14     len=0;

15     for(i=2;i<=2000009;i++)

16     {

17         if(s[i]==true) continue;

18         prime[++len]=i;

19         if(i>2000001) return;

20         for(j=i*2;j<=2000009;j=j+i)

21         s[j]=true;

22     }

23 }

24 int get_num(int n,int m){

25     int ans=0;

26     while(n){

27         n=n/m;

28         ans=ans+n;

29     }

30     return ans;

31 }

32 LL pow_mod(LL a,LL b)

33 {

34     LL ans=1;

35     while(b)

36     {

37         if(b&1){

38             ans=(ans*a)%mod;

39         }

40         b=b>>1;

41         a=(a*a)%mod;

42     }

43     return ans;

44 }

45 

46 void solve(int n,int m){

47     int ans;

48     LL sum1=1,sum2=1;

49     for(int i=1;i<=len;i++)

50     {

51         if(prime[i]>n+m)break;

52         ans=get_num(n+m,prime[i])-get_num(n,prime[i])-get_num(m,prime[i]);

53         sum1=(sum1*pow_mod(prime[i],ans))%mod;

54 

55         ans=get_num(n+m,prime[i])-get_num(n+1,prime[i])-get_num(m-1,prime[i]);

56         sum2=(sum2*pow_mod(prime[i],ans))%mod;

57     }

58    // printf("%lld %lld\n",sum1,sum2);

59     if(sum1<sum2) sum1=sum1-sum2+mod;

60     else sum1=sum1-sum2;

61     printf("%lld\n",sum1);

62 }

63 int main()

64 {

65     Init();

66     int T,n,m;

67     scanf("%d",&T);

68     while(T--){

69         scanf("%d%d",&n,&m);

70         solve(n,m);

71     }

72     return 0;

73 }

View Code