UVA10791

题目

UVA10791

做法

分解一下:\(n=p_1^{k_1}p_2^{k_2}...p_n^{k_n}\),易证\(a+b<a\cdot b(a,b≥2)\),把\(n\)分解成\(p_1^{k_1},p_2^{k_2}...,p_n^{k_n}\),有几个特判,细节看代码吧

My complete code

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
LL kase;
inline void Solve(LL n){
    LL ans(0),N(n),cnt(0);
    if(n==1){
        printf("Case %d: 2\n",++kase);
        return;
    }
    for(LL i=2;i*i<=N;++i){
        LL num(1);
        if(n%i==0 && n!=1){
            while(n%i==0 && n!=1){
                num*=i,
                n/=i;
            }
            ++cnt;
            ans+=num;
        }
    }
    if(n==N || cnt==1)
        ++ans;
    if(n!=1) ans+=n;
    printf("Case %d: %lld\n",++kase,ans);
}
int main(){
    LL n;
    while(scanf("%lld",&n)!=EOF&&n)
        Solve(n);
    return 0;
}

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转载自www.cnblogs.com/y2823774827y/p/10350896.html