Dice Cup

题目链接：

http://acm.hust.edu.cn/vjudge/contest/127406#problem/D

Description

In many table-top games it is common to use different dice to simulate random events. A “d” or “D”
is used to indicate a die with a specific number of faces, d4 indicating a four-sided die, for example.
If several dice of the same type are to be rolled, this is indicated by a leading number specifying the
number of dice. Hence, 2d6 means the player should roll two six-sided dice and sum the result face
values.
Write a program to compute the most likely outcomes for the sum of two dice rolls. Assume each
die has numbered faces starting at 1 and that each face has equal roll probability.

Input

The input file contains several test cases, each of them as described below.
The input consists of a single line with two integer numbers, N, M, specifying the number of faces
of the two dice.
Constraints:
4 ≤ N, M ≤ 20 Number of faces.

Output

For each test case, a line with the most likely outcome for the sum; in case of several outcomes with
the same probability, they must be listed from lowest to highest value in separate lines.
The outputs of two consecutive cases will be separated by a blank line.

Sample Input

6 6
6 4
12 20

Sample Output

7
5
6
7
13
14
15
16
17
18
19
20
21


##题意：

分别掷一个N面和M面的骰子.
求出现概率最大的顶面数和.


##题解：

直接模拟一遍记录所有顶面数之和.
把出现次数最多的数从小到大输出.


##代码：
``` cpp
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define LL long long
#define eps 1e-8
#define maxn 101000
#define mod 100000007
#define inf 0x3f3f3f3f
#define mid(a,b) ((a+b)>>1)
#define IN freopen("in.txt","r",stdin);
using namespace std;

int cnt[100];

int main(int argc, char const *argv[])

{

//IN;

int n, m;

int flag = 0;

while(scanf("%d %d", &n,&m) != EOF)

{

    memset(cnt, 0, sizeof(cnt));

    for(int i=1; i<=n; i++) {

        for(int j=1; j<=m; j++) {

            cnt[i+j]++;

        }

    }

    if(flag) printf("\n");

    flag = 1;

    int ma = 0;

    for(int i=0; i<=n+m; i++) ma = max(ma, cnt[i]);

    for(int i=0; i<=n+m; i++) {

        if(cnt[i] == ma)

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

    }

}

return 0;

}