1272: 求a＋aa＋aaa＋aa...a

Time Limit: 1 SecMemory Limit: 128 MB1272: 求a＋aa＋aaa＋aa...a

Submit: 1494Solved: 959

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Description

输入一个正整数 repeat (0

输入 2 个正整数a 和n, 求a+aa+aaa+aa⋯a(n 个a)之和。

Input

见sample

Output

见sample

Sample Input

1

2 3

Sample Output

246

---

思路：两个循环，一个用来求和一个用来求单个项

---

#include<stdio.h>

#include<iostream>

#include<math.h>

using namespace std;

int main(){

int i,j,a,n,sum,prom;

int repeat;

scanf("%d",&repeat);

while(repeat--)

{

scanf("%d%d",&a,&n);

sum=0;

prom=0;

for(i=0;i<n;i++)

for(j=0;j<=i;j++)

sum+=a*pow(10,j);

prom+=sum;

printf("%d\n",prom);

}

return 0;

}