HDU1868 Consecutive sum【数学计算+枚举】

Consecutive sum

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1590    Accepted Submission(s): 736

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

Sample Output

3 11

Author

Wiskey

Source

2008杭电集训队选拔赛——热身赛

 

 

 

问题链接：HDU1868 Consecutive sum。

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 */

#include <stdio.h>

int 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 */

#include <stdio.h>

int 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 <iostream>
#include <stdio.h>

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;
}