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 ).
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
Recommend
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 ",sum1,sum2); 59 if(sum1<sum2) sum1=sum1-sum2+mod; 60 else sum1=sum1-sum2; 61 printf("%lld ",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 }