UVALive 7275 Dice Cup （水题）

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("
");
    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
", i);
    }
}

return 0;

}