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


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\n", 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", n,m,ans);
        }
    }

    return 0;
}