2-06. Sum of series (20) (zjupat mathematics)

Link: http://pat.zju.edu.cn/contests/ds/2-06

Given a number a (1 <= A <= 9) and a non-negative integer N (0 <= n <= 100000), calculate the sum of the series S = a + AA + AAA +... + AA... A (n ). For example, if a = 1, n = 3, S = 1 + 11 + 111 = 123.

Input format description:

Enter the number a and the non-negative integer n.

Output format description:

Output the value of S in the sum of n columns.

Sample input and output:

Serial number	Input	Output
1	1 3	123
2	6 100	7407407407407407407407407407407407407407407407407407407407407407407407407407407407407407407407407340
3	1 0	0

The Code is as follows:

#include <cstdio>#include <cmath>int a[100017];int main(){    int A, N;    while(~scanf("%d%d",&A,&N))    {        if(N == 0)        {            printf("0\n");            continue;        }        int tt = 0,p = 0;        int j = 0;        for(int i = N; i >= 1; i--)        {            tt = A*i+p;            p = tt/10;            a[j++] = tt%10;        }        if(p > 0)        {            a[j++] =  p;        }        for(int i = j-1; i >= 0; i--)        {            printf("%d",a[i]);        }        printf("\n");    }    return 0;}

2-06. Sum of series (20) (zjupat mathematics)