CS Academy Consecutive Sum

题目链接：https://csacademy.com/contest/archive/task/consecutive-sum

题目大意：给出一个数n，判断它是否能够被至少两个的连续整数和表示出来。能的话输出A到B，A+A+1+……+B = n，否则输出-1。

解题思路：使用双指针，从前到后计算前i个数的和，如果这个数超过n，就从j=1开始减去j，如果减法过程中当前和等于n，直接退出，如果小于，继续向前将下一个数加入当前和中继续判断。

代码：

const int maxn = 1e6 + 5;
int n;

void solve(){
    int i = 1, j = 1;
    ll tmp = 0;
    for(i; i <= n; i++){
        tmp += i;
        if(tmp < n) continue;
        else if(tmp == n) break;
        else{
            while(tmp > n){
                tmp -= j;
                j++;
            }
            if(tmp == n) break;
        }
    }
    if(j < i && tmp == n) printf("%d %d
", j, i);
    else puts("-1");
    return;
} 
int main(){
    scanf("%d", &n);
    solve();
}

题目：

Consecutive Sum

Time limit: 1000 ms
Memory limit: 256 MB

 

You are given a number NN. Write NN as a sum of at least 22 positive consecutive integers.

Standard input

The first line contains a single integer NN.

Standard output

Print two integers AA and BB, representing the smallest and the largest terms. Basically, NN should be equal to A + (A + 1) + ... + BA+(A+1)+...+B. If there are multiple solutions you can output any of them.

Constraints and notes

• 1 leq N leq 10^51≤N≤10​5​​ 
• 1 leq A < B1≤A<B

Input	Output	Explanation
7	3 4	3+4=73+4=7
15	1 5	1+2+3+4+5=151+2+3+4+5=15
4	-1