SPOJ - LOCKER 数论 贪心

题意:求出\(n\)拆分成若干个数使其连乘最大的值

本题是之江学院网络赛的原题,计算规模大一点,看到EMAXX推荐就做了

忘了大一那会是怎么用均值不等式推出结果的(还给老师系列)

结论倒还记得:贪心分解3,不够就用2凑

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define rrep(i,j,k) for(register int i=j;i>=k;i--)
#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])
#define iin(a) scanf("%d",&a)
#define lin(a) scanf("%lld",&a)
#define din(a) scanf("%lf",&a)
#define s0(a) scanf("%s",a)
#define s1(a) scanf("%s",a+1)
#define print(a) printf("%lld",(ll)a)
#define enter putchar(‘\n‘)
#define blank putchar(‘ ‘)
#define println(a) printf("%lld\n",(ll)a)
#define IOS ios::sync_with_stdio(0)
using namespace std;
const int maxn = 1e6+11;
const int oo = 0x3f3f3f3f;
const double eps = 1e-7;
typedef long long ll;
ll read(){
    ll x=0,f=1;register char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
ll fpw(ll a,ll n,ll mod){
    ll ans=1;
    while(n){
        if(n&1) ans=(ans*a)%mod;
        n>>=1;
        a=(a*a)%mod;
    }
    return ans;
}
int main(){
    int T=read();
    const int mod = 1e9+7;
    while(T--){
        ll n=read();
        ll ans=0;
        if(n%3==0){
            ans=fpw(3,n/3,mod);
        }else if(n%3==1){
            if(n==1)ans=1;
            else ans=fpw(3,n/3-1,mod)*4;//3 1 -> 2 2
        }else{
            if(n==2)ans=2;
            else ans=fpw(3,n/3,mod)*2;//3 2
        }
        println((ans%mod));
    }
    return 0;
}

原文地址：https://www.cnblogs.com/caturra/p/8457025.html