HDU 5793 A Boring Question 打表找规律

打个表就会发现这是很有规律的 答案为( m^（n+1）-1)/(m-1)

ACcode:

#pragma warning(disable:4786)//使命名长度不受限制
#pragma comment(linker, "/STACK:102400000,102400000")//手工开栈
#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <stack>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
#define rd3(x,y,z) scanf("%d%d%d,&x,&y,&z)
#define rdl(x) scanf("%I64d,&x);
#define rds(x) scanf("%s",x)
#define rdc(x) scanf("%c",&x)
#define ll long long int
#define ull unsigned long long
#define maxn 1005
#define mod 1000000007
#define INF 0x3f3f3f3f //int 最大值
#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)
#define MT(x,i) memset(x,i,sizeof(x))
#define PI acos(-1.0)
#define E exp(1)
#define eps 1e-8
ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}
ll pow_(ll a, ll n){
ll ans=1;
while(n){
if(n&1)ans=(ans*a)%mod;
a=(a*a)%mod;
n>>=1;
}
return ans;
}
ll inv(ll a){if(a==1)return 1; return inv(mod%a)*(mod-mod/a)%mod;}
using namespace std;
int k[maxn];
int N(int x){
if(x==0||x==1)return 1;
int res=1;
for(int i=1;i<=x;++i)res*=i;
return res;
}
int fun(int j,int i){
if(j<i)return 0;
else return N(j)/(N(i)*(N(j-i)));
}
int doit(int a,int b){
int res=0;
if(b==2){
for(int i=0;i<=a;++i)
for(int j=0;j<=a;++j)
res=(res+fun(j,i))%mod;
}
if(b==3){
for(int i=0;i<=a;++i)
for(int j=0;j<=a;++j)
for(int z=0;z<=a;++z)
res=(res+fun(j,i)*fun(z,j))%mod;
}
if(b==4){
for(int k1=0;k1<=a;++k1)
for(int k2=0;k2<=a;++k2)
for(int k3=0;k3<=a;++k3)
for(int k4=0;k4<=a;++k4)
res=(res+fun(k4,k3)*fun(k3,k2)*fun(k2,k1))%mod;
}
if(b==5){
for(int k1=0;k1<=a;++k1)
for(int k2=0;k2<=a;++k2)
for(int k3=0;k3<=a;++k3)
for(int k4=0;k4<=a;++k4)
for(int k5=0;k5<=a;++k5)
res=(res+fun(k5,k4)*fun(k4,k3)*fun(k3,k2)*fun(k2,k1))%mod;
}
if(b==6){
for(int k1=0;k1<=a;++k1)
for(int k2=0;k2<=a;++k2)
for(int k3=0;k3<=a;++k3)
for(int k4=0;k4<=a;++k4)
for(int k5=0;k5<=a;++k5)
for(int k6=0;k6<=a;++k6)
res=(res+fun(k6,k5)*fun(k5,k4)*fun(k4,k3)*fun(k3,k2)*fun(k2,k1))%mod;
}
if(b==7){
for(int k1=0;k1<=a;++k1)
for(int k2=0;k2<=a;++k2)
for(int k3=0;k3<=a;++k3)
for(int k4=0;k4<=a;++k4)
for(int k5=0;k5<=a;++k5)
for(int k6=0;k6<=a;++k6)
for(int k7=0;k7<=a;++k7)
res=(res+fun(k7,k6)*fun(k6,k5)*fun(k5,k4)*fun(k4,k3)*fun(k3,k2)*fun(k2,k1))%mod;
}
return res;
}
int main(){
int n,cnt=1;
// for(int i=1;i<=10;++i)
// for(int j=2;j<=7;++j)
// cout<<i<<" and "<<j<<" is "<<doit(i,j)<<"\12";
ll a,b,loop;
scanf("%I64d",&loop);
while(loop--){
scanf("%I64d%I64d",&a,&b);
int ans=(pow_(b,a+1)-1)*inv(b-1)%mod;
cout<<ans<<'\12';
}
return 0;
}