HDU 5793 A Boring Question （找规律 : 快速幂+逆元）

A Boring Question

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5793

Description

![](http://images2015.cnblogs.com/blog/764119/201608/764119-20160804173305559-605626236.png)

Input

The first line of the input contains the only integer T, Then T lines follow,the i-th line contains two integers n,m.

Output

For each n and m,output the answer in a single line.

Sample Input

2 1 2 2 3

Sample Output

3 13

Source

2016 Multi-University Training Contest 6
##题意：用m个不大于n的数构成一个序列，对每个序列求C(ki+1,ki)的连乘积. 对所有可能的序列，累加上述连乘积.
##题解：还是打表找的规律...(好弱啊) f(1,2)=3; f(2,2)=7; f(1,3)=4; f(2,3)=13; f(1,4)=5; f(2,4)=21; f(1,5)=6; f(2,5)=31; ...... 打了个5*5的表后发现规律：(后附打表代码) f(n,m) = f(n-1,m) + m^n; = m^0 + m^1 + m^2 + ... + m^n; (等比数列求和) = (1 - m^(n+1)) / (1 - m); 然后用快速幂和乘法逆元求出上式即可.

官方题解：

##代码： ``` cpp #include #include #include #include #include #include #include #include #include #define LL long long #define mid(a,b) ((a+b)>>1) #define eps 1e-8 #define maxn 2100 #define mod 1000000007 #define inf 0x3f3f3f3f #define IN freopen("in.txt","r",stdin); using namespace std;

LL x,y,gcd;
void ex_gcd(LL a,LL b)
{
if(!b) {x=1;y=0;gcd=a;}
else {ex_gcd(b,a%b);LL temp=x;x=y;y=temp-a/b*y;}
}

LL quickmod(LL a,LL b,LL m) {
LL ans = 1;
while(b){
if(b&1){
ans = (ansa)%m;
b--;
}
b/=2;
a = aa%m;
}
return ans;
}

int main(int argc, char const *argv[])
{
//IN;

int t; cin >> t;
LL n, m;
while(scanf("%I64d %I64d", &n,&m) != EOF)
{
    LL ans1 = quickmod(m, n+1, 1000000007LL) - 1;
    LL ans2 = m - 1;

    ex_gcd(ans2, 1000000007LL);
    while(x < 0) {
        x+=1000000007LL;
        y-=ans2;
    }

    LL ans = (ans1 * x) % mod;

    printf("%I64d
", ans);
}

return 0;

}


####打表代码：
``` cpp
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <vector>
#define LL long long
#define mid(a,b) ((a+b)>>1)
#define eps 1e-8
#define maxn 2100
#define mod 1000000007
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;

LL e[510][510];
void make(){
    for(int i=0;i<510;i++)
        e[i][0]=1;
    for(int i=1;i<510;i++)
        for(int j=1;j<510;j++)
            e[i][j]=(e[i-1][j-1]+e[i-1][j])%mod;
}

int n,m,ans;
void fun(int len, vector<int> cur, int last) {
    if(len == m) {
        int tmp = 1;
        for(int i=1; i<cur.size(); i++) {
            tmp *= e[cur[i]][cur[i-1]];
        }
        ans += tmp;
        return;
    }

    for(int i=last; i<=n; i++) {
        cur.push_back(i);
        fun(len+1, cur, i);
        cur.pop_back();
    }
}

int main(int argc, char const *argv[])
{
    //IN;

    make();

    for(n=0; n<=5; n++) {
        for(m=2; m<=5; m++) {
            ans = 0;
            vector<int> cur; cur.clear();
            fun(0,cur,0);
            printf("%d-%d : %d
", n,m,ans);
        }
    }

    return 0;
}

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原文地址：https://www.cnblogs.com/Sunshine-tcf/p/5737627.html