WHUST 2016 Summer Contest #4D-Dice Cup

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

题目大意

分别投一个有N面的骰子和一个有M面的的骰子，求出现概率最大的点数之和

解题思路

思路一：由于数据范围不大，所以可以直接记录所有的点数之和出现的次数，然后输出出现次数最多的点数和 思路二：打表找规律，通过打表，我们发现，点数和出现的最多次数不会超过N和M的最小值，所以，只用输出出现次数大于min（N，M）的点                   数和即可 思路三：观察样例，我们发现6+1<=7<=6+1;4+1<=5,6,7<=6+1;……所以，输出的数其实就是N+1到M+1间的所有数（N+1和M+1也要输出）

注意事项

只有两个连续的cases之间才要用空行隔开，所以最后一个case结束之后不需要空行

【代码】

#include <iostream> using namespace std; int main() { int n,m,f=1; while(cin>>n>>m) { if(f) f=0; else cout<<endl; for(int i=min(n,m)+1;i<=max(n,m)+1;i++) cout<<i<<endl; } return 0; }