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.

 

 

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).

 

 

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;
}