LightOJ 1278 - Sum of Consecutive Integers 分解奇因子 + 思维

http://www.lightoj.com/volume_showproblem.php?problem=1278

题意：问一个数n能表示成几种连续整数相加的形式如6=1+2+3，1种。

思路：先列式子\(N=a+(a+1)+(a+2)+ ...+(a+k-1)=\frac{k·(2a+k-1)}{2} \) 继续化成\(2a-1=\frac{2N}{k} - k \) 可由左式得知，2a-1必为奇数，那么右式必定是一奇一偶，且都为2N的因子。所以只要分解因子记录个数，最后组合求一下即可。

/** @Date    : 2016-11-24-22.15* @Author  : Lweleth (SoungEarlf@gmail.com)* @Link    : https://github.com/* @Version :*/
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <utility>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <stack>
#include <queue>
//#include<bits/stdc++.h>
#define LL long long
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;const int INF = 0x3f3f3f3f;
const int N = 1e6+20;
int pri[670000];
int c = 0;
bool vis[N*10];void prime()
{MMF(vis);for(int i = 2; i < N*10; i++){if(!vis[i])pri[c++] = i;for(int j = 0; j < c && i * pri[j] < N*10; j++){vis[i*pri[j]] = 1;if(i % pri[j] == 0)break;}}
}int main()
{int T;int cnt = 0;cin >> T;prime();while(T--){LL n;scanf("%lld", &n);LL ans = 1;//while(n % 2 == 0)//    n/=2;for(int i = 0; i < c && pri[i]*pri[i] <= n; i++)//记录素因子为奇数的（除2外的） {LL ct = 0;while(n % pri[i] == 0){if(i != 0)ct++;n /= pri[i];}ans *= (ct + 1);}if(n > 1 && n != 2){ans *= 2;}printf("Case %d: %lld\n", ++cnt, ans-1);}return 0;
}

转载于:https://www.cnblogs.com/Yumesenya/p/6123939.html

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