Problem Description
Every body knew that 15 = 1+2+3+4+5 = 4+5+6 = 7+8. Now give you a number N, tell me how many ways to represent N as a sum of consecutive positive integers. For example, 15 have 3 ways to be found.
Input Each line will contain an signed 32-bits integer N. Process to end of file.
Output For each case, output the answer in one line.
Sample Input 15 1050
3 11
Wiskey
Source 问题链接：。
题意简述：参见上文。
问题分析：这个问题还是需要用数学方法来解决，即先做数学推导再编写暴力程序，否则容易TLE。
对于输入的n，假设首项和末项为a和b的序列其和为n，即(a+b)*(b-a+1)/2==n。
设a+b=x,b-a+1=y,则有方程组 x*y=n*2, 两式相加得x+y=2*b+1，故x+(2*n/x)=2*b+1。
所以只要检测x能否整除2*n，并且使上面方程满足中b为正整数的情况。
程序说明：（略）
题记：（略）
AC的C语言程序如下：
/* HDU1868 Consecutive sum */#includeint main(void){ int n, ans, i; while(scanf("%d", &n) != EOF) { n*=2; ans = 0; for(i=2; i*i<=n; i++) if(n % i==0 && (i + n / i) % 2 == 1) ans++; printf("%d\n", ans); } return 0;}
另外一个AC的C语言程序如下：
/* HDU1868 Consecutive sum */#includeint main(void){ int n, ans, i; while(scanf("%d", &n) != EOF) { ans = 0; if(n % 2) ans++; for(i=2; i*i<=n; i++) if(n % i == 0) { if(i % 2 != 0) ans++; if(n / i % 2 != 0) ans++; } printf("%d\n", ans); } return 0;}
TLE的C++语言程序（尺取法）如下：
/* HDU1868 Consecutive sum */#include#include using namespace std;int main(){ int n; while(scanf("%d", &n) != EOF) { int start, end, sum, ans; start = end = 1; sum = ans = 0; for(;;) { while(end <= n && sum < n) sum += end++; if (sum == n && end - start != 1) ans++; sum -= start++; if (sum <= 0) break; } printf("%d\n", ans); } return 0;}