URAL 1264 Workdays （水）

Workdays

Time limit: 1.0 second
Memory limit: 64 MB

After a success of the previous Vasechkin’s program that allowed to calculate the results of the elections in cause of two days Artemy Sidorovich was placed at the head of the department. At the moment Artemy Sidorovich prepares a task for his subordinate — programmer Petechkin. The task is to write a very useful function that would ease the life of all the department programmers. For each integer from 0 to M the function would calculate how many times this number appears in the N-element array. Artemy Sidorovich deems that the function should work as follows (the sample code for N = 3,M = 1):

CPascal

if (arr[0]==0) ++count[0];if (arr[0]==1) ++count[1];if (arr[1]==0) ++count[0];if (arr[1]==1) ++count[1];if (arr[2]==0) ++count[0];if (arr[2]==1) ++count[1];

if arr[0]=0 then count[0] := count[0] + 1;if arr[0]=1 then count[1] := count[1] + 1;if arr[1]=0 then count[0] := count[0] + 1;if arr[1]=1 then count[1] := count[1] + 1;if arr[2]=0 then count[0] := count[0] + 1;if arr[2]=1 then count[1] := count[1] + 1;

Artemy Sidorovich wants to estimate the time that Petechkin will need to execute the task. We know that Petechkin needs one second to write a line of the code (he’s fast, isn’t he?). Artemy Sidorovich doesn’t know exactly bounds for M and N. Your task is to write program that would calculate a number of seconds that Petechkin will write the code.

Input

The only line contains integers N (0 ≤ N ≤ 40000) and M (0 ≤ M ≤ 40000).

Output

Output an amount of seconds that Petechkin needs to write the program.

Sample

inputoutput

3 1

6

解析：m从0开始，所以结果是n * (m + 1)。

AC代码：

#include <cstdio>int main(){    int n, m;    while(~scanf("%d%d", &n, &m)){        printf("%d\n", n * (m + 1));    }    return 0;}

PS：读完题，好久没敢写。。。竟然就是这样。。。