LightOJ 1278 Sum of Consecutive Integers

题目分析

这道题一开始没有思路，然后看了网上的推导过程。具体是这样：sum=a+(a+1)+(a+2)+......+(a−k+1),然后可以推出式子n=(2a+k−1)/(2k),2a−1=2n/k−k，因为2a−1为奇数，所以k必须为奇数。（奇数*偶数=偶数）,然后直接求n的奇素因子即可。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define LL long long
const int maxn = 1e7+100;

bool vis[maxn];
int prime[maxn/10];

void init(){
memset(vis, false, sizeof(vis));
int tot = 0;
for(int i = 2; i < maxn; i++)if(!vis[i]){
prime[tot++] = i;
for(LL j = (LL)i*i; j < maxn; j += i) vis[j] = true;
}
}

int main(){
init();
int T;
LL n;
scanf("%d", &T);
for(int kase = 1; kase <= T; kase++){
scanf("%lld", &n);
while(n%2 == 0) n /= 2;
int ans = 1;
for(int i = 0; (LL)prime[i]*prime[i] <= n; i++)if(n%prime[i] == 0){
int tot = 1;
while(n%prime[i] == 0){
tot++;
n /= prime[i];
}
ans *= tot;
}
if(n != 1) ans *= 2;
printf("Case %d: %d\n", kase, ans-1);
}
return 0;
}