【HDOJ5793】A Boring Question（数学题）

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记录一个菜逼的成长。。

找规律推导。
ans = (m^(n+1) - 1) / (m-1)
利用快速幂和乘法逆元对分数取模

#include <cstdio>#include <iostream>#include <cstring>#include <string>#include <algorithm>#include <cstdlib>#include <vector>#include <set>#include <map>#include <queue>#include <list>#include <deque>#include <cctype>#include <bitset>using namespace std;#define cl(a) memset(a,0,sizeof(a))typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> PII;const int INF = 0x3f3f3f3f;const LL MOD = 1000000007;LL PowerMod(LL a, LL b, LL c){    ULL ans = 1;    a = a % c;    while(b>0)    {        if(b % 2 == 1)            ans = (ans * a) % c;        b = b / 2;        a = (a * a) % c;    }    return ans;}LL gcd(LL a,LL b){    return b==0?a:gcd(b,a%b);}void exgcd(LL a,LL b,LL &x,LL &y){    if(b==0)    {        x=1;        y=0;        return ;    }    exgcd(b,a%b,x,y);    LL t=x;    x=y;    y=t-(a/b)*y;}int main(){    int T;    scanf("%d",&T);    while(T--){        LL n,m;        scanf("%lld%lld",&n,&m);        LL x, y;        LL tmp = PowerMod(m,n+1,MOD) - 1,mod = MOD;        m -= 1;        LL Gcd = __gcd(m,mod);        m /= Gcd;mod /= Gcd;        exgcd(m,mod,x,y);        LL ans = (x%mod+mod)%mod;        printf("%lld\n",tmp * ans % MOD);    }    return 0;}

或者

#include<bits/stdc++.h>#define mod 1000000007using namespace std;long long powt(long long a,long long b){    long long r = 1;    while(b)    {        if(b & 1) r = r * a % mod;        a = a * a % mod;        b >>= 1;    }    return r;}int main(){    long long t,n,m;    scanf("%lld",&t);    while(t--)    {        scanf("%lld%lld",&n,&m);        printf("%lld\n",((powt(m,n + 1) - 1) * powt(m - 1,mod - 2) % mod + mod) % mod);    }    return 0;}

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