hdu 多校
Problem B. Harvest of Apples
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1006 Accepted Submission(s): 381
Problem Description
There are n apples on a tree, numbered from 1 to n.
Count the number of ways to pick at most m apples.
Count the number of ways to pick at most m apples.
Input
The first line of the input contains an integer T (1≤T≤105) denoting the number of test cases.
Each test case consists of one line with two integers n,m (1≤m≤n≤105).
Each test case consists of one line with two integers n,m (1≤m≤n≤105).
Output
For each test case, print an integer representing the number of ways modulo 109+7.
Sample Input
2
5 2
1000 500
Sample Output
16
924129523
题意
t次询问,求C(n,0)+C(n,1)+...C(n,m)
思路
通过暴力打表 100000x100000 这样会超时 ,因为C(n,m)=C(n-1,m-1)+C(n-1,m),所以sum[n][m]=2*sum[n-1][m]-C(n-1,m)
这样就可以通过某个点求出其他点的值,用分块,每200x200的矩阵压成一个点,这样就可以快速查询。
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; const int mod=1e9+7; typedef long long ll; int len=200; ll A[100005],p[100005],q[100005]; ll dp[505][505]; ll qpow(ll a,ll b,ll c) { ll ans=1; while(b) { if(b&1) ans=ans*a%c; a=a*a%c; b>>=1; } return ans; } ll C(int n,int m) { return (A[n]*p[m])%mod*p[n-m]%mod; } void init() { ll sum; for(int i=0;i<=500;i++) { int m=i*len; sum=0; for(int j=0;j<=m;j++) { sum+=C(m,j); sum%=mod; if(j%len==0) { dp[i][j/len]=sum; } } } } ll get(int n,int m) { int x,y; x=n/len*len; y=m/len*len; ll ans; ans=dp[n/len][m/len]; for(int i=x;i<n;i++) { ans=(ans*2-C(i,y)); ans=(ans+mod)%mod; } for(int i=y+1;i<=m;i++) { ans+=C(n,i); ans=ans%mod; } return ans; } int main() { int i,j,k,t,n,m; A[0]=1; p[0]=1; q[0]=1; for(i=1; i<=100000; i++) { A[i]=A[i-1]*i%mod; p[i]=qpow(A[i],mod-2,mod); q[i]=2*q[i-1]%mod; } init(); scanf("%d",&t); ll ans; while(t--) { scanf("%d%d",&n,&m); printf("%I64d\n",get(n,m)); } return 0; }