2014
02-12

# Coin

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.

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)

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)

1
5 7

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

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。”