首页 > ACM题库 > HDU-杭电 > hdu 2662 Coin-最小生成树-[解题报告]C++
2014
02-12

hdu 2662 Coin-最小生成树-[解题报告]C++

Coin

问题描述 :

Moon has many coins, but only contains two value types which is 5 cents and 7 cents, Some day he find that he can get any value which greater than 23 cents using some of his coins. For instance, he can get 24 cents using two 5 cents coins and two 7 cents coins, he can get 25 cents using five 5 cents coins, he can get 26 cents using one 5 cents coins and three 7 cents coins and so on.

Now, give you many coins which just contains two value types just like Moon, and the two value types identified by two different prime number i and j. Can you caculate the integer n that any value greater than n can be created by some of the given coins.

输入:

The first line contains an integer T, indicates the number of test cases.
For each test case, there are two different prime i and j separated by a single space.(2<=i<=1000000, 2<=j<=1000000)

输出:

The first line contains an integer T, indicates the number of test cases.
For each test case, there are two different prime i and j separated by a single space.(2<=i<=1000000, 2<=j<=1000000)

样例输入:

1 
5 7

样例输出:

23

#include <iostream>
 using namespace std;
 
 typedef unsigned long long int longint;
 
 
 int main()
 {
     int t;
     cin >> t;
     while (t--)
     {
         longint n, m;
         cin >> n >> m;
         longint ans;
 
         ans = (n * m) - n - m;
 
         cout << ans << endl;
     }
     return 0;
 }
 
 // end
 // ism

解题转自:http://www.cnblogs.com/izumu/archive/2012/07/20/2600867.html


  1. 有一点问题。。后面动态规划的程序中
    int dp[n+1][W+1];
    会报错 提示表达式必须含有常量值。该怎么修改呢。。

  2. 5.1处,反了;“上一个操作符的优先级比操作符ch的优先级大,或栈是空的就入栈。”如代码所述,应为“上一个操作符的优先级比操作符ch的优先级小,或栈是空的就入栈。”