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：

 1 const int maxn = 1e6 + 5;
 2 int n;
 3 
 4 void solve(){
 5     int i = 1, j = 1;
 6     ll tmp = 0;
 7     for(i; i <= n; i++){
 8         tmp += i;
 9         if(tmp < n) continue;
10         else if(tmp == n) break;
11         else{
12             while(tmp > n){
13                 tmp -= j;
14                 j++;
15             }
16             if(tmp == n) break;
17         }
18     }
19     if(j < i && tmp == n) printf("%d %d\n", j, i);
20     else puts("-1");
21     return;
22 } 
23 int main(){
24     scanf("%d", &n);
25     solve();
26 }

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

 

InputOutputExplanation
 
7

3 4

3+4=73+4=7

 
15

1 5

1+2+3+4+5=151+2+3+4+5=15

 
4

-1
posted @ 2017-09-10 09:57  EricJeffrey  阅读(187)  评论(0)  编辑  收藏  举报