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最新推荐文章于 2025-08-23 19:36:46 发布

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原文链接：http://www.cnblogs.com/liweiggg/p/9405329.html

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#java

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

Input

The first line of the input contains an integer

Output

For each test case, print an integer representing the number of ways modulo

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;
}