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  • 杭电acm5698-瞬间移动(2016"百度之星"

    题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=5698

    Problem Description

    有一个无限大的矩形,初始时你在左上角(即第一行第一列),每次你都可以选择一个右下方格子,并瞬移过去(如从下图中的红色格子能直接瞬移到蓝色格子),求到第nnn行第mmm列的格子有几种方案,答案对100000000710000000071000000007取模。

    http://acm.hdu.edu.cn/data/images/C702-1003-1.jpg

    Input

    多组测试数据。

    两个整数n,m(2≤n,m≤100000)n,m(2leq n,mleq 100000)n,m(2n,m100000)

    Output

    一个整数表示答案

    Sample Input
    4 5
    Sample Output
    10
     
     
     
     
    /*
    #include<stdio.h>
    #include<stdlib.h>
    int a[300][300]={0};
    int main()
    {
        int n,m,i,j; 
        //printf("%d
    ",a[1][1]);
        for (i=2;i<300;i++)
        {
            a[i][2] = 1;
            a[2][i] = 1;
        }
        for (i=3;i<300;i++)
        {
            for (j=3;j<300;j++)
            {
                a[i][j] = a[i][j-1]+a[i-1][j];
                a[i][j] = a[i][j]%1000000007;
            }
        }
        while (scanf("%d%d",&n,&m)!=EOF)
        {
            printf("%d
    ",a[n][m]);
        }
        return 0;
    }
    */
    
    //ac代码
    #include<iostream>
    #include<cstring>
    #include<cstdio>
    #include<queue>
    #include<stack>
    #include<map>
    #define ll __int64
    #define mod 1000000007
    using namespace std;
    ll poww(ll a,ll n)
    {
       ll r=1,p=a;
       while(n)
       {
           if(n&1) r=(r*p)%mod;
           n>>=1;
           p=(p*p)%mod;
       }
       return r;
    }
    ll coun(ll n,ll m) 
    {
        ll sum=1; 
        for(ll i=1,j=n;i<=m;i++,j--)
        {
            sum*=j;
            sum%=mod;
            sum*=poww(i,1000000005);
            sum%=mod;
        }
        return sum;
    }
     ll x,y,z,t; 
     ll n,k;
     ll ans;
    int main()
    {
        while(scanf("%I64d %I64d",&x,&y)!=EOF)
        {
            ans=0;
            n=(x+y)-4;
            k=x-2;
            ans=coun(n,k);
            cout<<ans<<endl;
        }
        return 0;
    }
    
    
    
    /*
    //ac代码
    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<string>
    #include<queue>
    #include<algorithm>
    #include<stack>
    #include<cstring>
    #include<vector>
    #include<list>
    #include<set>
    #include<map>
    using namespace std;
    #define ll long long
    #define mod 1000000007
    #define inf 999999999
    #define pi 4*atan(1)
    //#pragma comment(linker, "/STACK:102400000,102400000")
    int scan()
    {
        int res = 0 , ch ;
        while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) )
        {
            if( ch == EOF ) return 1 << 30 ;
        }
        res = ch - '0' ;
        while( ( ch = getchar() ) >= '0' && ch <= '9' )
            res = res * 10 + ( ch - '0' ) ;
        return res ;
    }
    void extend_Euclid(ll a, ll b, ll &x, ll &y)
    {
        if(b == 0)
        {
            x = 1;
            y = 0;
            return;
        }
        extend_Euclid(b, a % b, x, y);
        ll tmp = x;
        x = y;
        y = tmp - (a / b) * y;
    }
    ll combine1(ll n,ll m) //计算组合数C(n,m)
    {
        ll sum=1; //线性计算
        for(ll i=1,j=n;i<=m;i++,j--)
        {
            sum*=j;
            sum%=mod;
            ll x,y;
            extend_Euclid(i,mod,x,y);
            sum*=(x%mod+mod)%mod;
            sum%=mod;
        }
        return sum;
    }
    int main()
    {
        ll x,y,z,i,t;
        while(~scanf("%I64d%I64d",&x,&y))
        {
            ll n,k;
            ll ans=0;
            k=x-2;
            n=(x+y)-4;
            ans=combine1(n,k);
            printf("%I64d
    ",ans);
        }
        return 0;
    }
    
    */
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  • 原文地址:https://www.cnblogs.com/acmwangpeng/p/5532797.html
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