编程题：输入一个正整数，若该数能用几个连续正整数之和表示，则

public class ConsecutiveIntegerSum {

    public static void main(String[] args) {
        devideIntoConsecutiveIntegerSum(10000);
    }
    
    public static void devideIntoConsecutiveIntegerSum(int num){
        int a1 = 1;
        int n = 2;
        for(n = 2 ; n < num ;n++){
            for(a1 = 1; a1 < num ;a1++){
                if((n*n + (2*a1 - 1)*n )  == 2 *num){
                    for(int i=1;i<=n;i++){
                        System.out.print((a1 + (i-1)) + ",");
                    }
                    System.out.println();
                }
            }
        }
    }
}

#include <stdio.h>
#include <malloc.h>

void devide (int *arr, int len, int num, int thr);
void devide_rec (int *arr, int len, int num, int thr);

void printArr (int *arr, int len) {
    int i = len-1;
    while (arr[i]) {
        printf ("%d ", arr[i]);
        i -= arr[i];
    }
    printf ("\n");
}

void devide (int *arr, int len, int num, int thr) {
    arr[num] = thr;
    num -= thr;
    if (!num) {
        printArr (arr, len);
        return;
    }
    devide_rec (arr, len, num, thr);
}

void devide_rec (int *arr, int len, int num, int thr) {
    int i;
    if (thr > num) {
        thr = num;
    }
    for (i = 1; i <= thr; ++i) {
        devide(arr, len, num, i);
    }
}

int main () {
    int num = 6;
    int *arr = (int*)malloc(sizeof(int)*(num+1));
    arr[0] = 0;
    devide_rec (arr, num+1, num, num);
    free(arr);
    return 0;
}

/*
设初始值为x，有m个数，因为是连续的数字，那么求和为mx + m*(m-1)/2 = n。
因为x至少为1，那么代入可推出m<=√(2*n)，而且m>=2。然后枚举m，由等式可求出x，
如果x为整数，那么x就满足题意，可输出。效率可达到o(√n)
*/

#include<iostream>
#include<cmath>

using namespace std;

void main()
{
    int n;

    cin >> n;

    for(int m = sqrt(2*n); m >=2 ; m --)
    {
        double fx = (double)(2*n - m*(m-1)) / (2*m);
        int dx = (int)fx;

        if(dx == fx)
        {
            for(int i = dx; i < dx+m; i ++)
            {
                cout << i << " ";
            }
            cout << endl;
        }
    }
}

int Fun(int n) {
    int m = (n+1)/2;
    for ( int i=2;i<=m;++i ) {
        if ( 0==(n-(i-1)*i/2)%i ) {
            return (n-(i-1)*i/2)/i);
        }
    }
    return -1;
}

import java.util.Scanner;

public class GetSum {
    public static void main(String[] argv){
        Scanner scanner = new Scanner(System.in);
        
        int n = scanner.nextInt();
        
        for(int i=1, startNum=1, sum=0;i<=n/2;i++){
            sum += i;
            System.out.println("i :" + i + " startNum :" + startNum + " sum :" + sum);
            if(sum > n){
                sum = 0;
                i = startNum;
                startNum+=1;
                
                continue;
            }
            
            if(sum == n){
                for(int j=startNum;j<=i;j++){
                    System.out.print(j + " ");
                }
                System.out.println();
                
                sum = 0;
                i = startNum;
                startNum += 1;
            }
        }
    }
}

	

int input;    //开始时输入的正整数，设定为全局变量

int main()
{
    int i;
    input = 9;    //赋值9，答案：4,5和2,3,4

    for (i = 1; i < input; i++)    //起始值设定为1开始往上升，直到input-1结束
         calResult(0, i);

    return 0;
}

int calResult(int sum, int num)
{
    int num_t;
    int sum_t;
    int ret=0;
    if (sum + num == input)    //如果跟input一致就判断找到连续和跟input一致的结果
    {
        printf("number: %d",num);
        return 1;
    }
    else if (sum + num > input)    //如果大于input返回-1终止递归
        return -1;
    else
    {
        sum_t = sum + num;
        num_t = num + 1;
        ret = calResult(sum_t, num_t);    //使用递归查看连续累加的数值跟input比较

        if (ret == 1)
            printf(",%d",num);
        return ret;
    }
}
             

int i=0,j=0,sum=0,nstart，nend;
bool flag = false;
for(i=1;i<n/2;i++)
{
    sum = 0;
    nsart = i;
    for(j=1;i<=i;j++)
        {
            sum+=j;
            if(sum == targert)
            {
                nstart = j；
                bool = true;
                break;
            }
        }
}
         

	#include<stdio.h>



	



	int main()



	{



	    int i,z,x,y,j;



	    printf("please input z:");



	    scanf("%d",&z);



	    for(i=1;i<=z;i++)



	        for(x=1;x<=z;x++)



	        {



	            y=x+i-1;



	            if((y+x)*i==2*z)



	            {



	                for(j=x;j<=y;j++)



	                    printf("%3d ",j);



	                printf("\n"); 



	            }



	        }



	}