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Problem Description：
Given three integers A, B and C in [−
,
], you are supposed to tell whether A B>C.
Input Specification:
The first line of the input gives the positive number of test cases, T (≤10). Then T test cases follow, each consists of a single line containing three integers A, B and C, separated by single spaces.
Output Specification:
For each test case, output in one line Case #X: true if A B>C, or Case #X: false otherwise, where X is the case number (starting from 1).
Sample Input:
代码语言：javascript复制3
1 2 3
2 3 4
9223372036854775807 -9223372036854775808 0Sample Output:
代码语言：javascript复制Case #1: false
Case #2: true
Case #3: false解题思路：
我一开始看到题目说A,B,C都在[-
,
]这个区间内时，只想到了要用long long型，并没有考虑到A B后会出现数据溢出的问题，所以有俩个测试用例WA，20分只得了12分。绝望的我去看了一下柳神的代码(1065. A B and C (64bit) (20)-PAT甲级真题)后发现这题会有俩种数据溢出的情况：令sum=A B，①当A>0且B>0时会出现数据溢出，数据溢出后sum是个负数，此时A B>C是成立的；②同理当A<0且B<0时也会出现数据溢出，数据溢出后sum是个正数，此时A B肯定是小于C的。这俩种情况分别是4分。。。。。
AC代码： 12分代码：
代码语言：javascript复制#include <bits/stdc .h>
using namespace std;
int main()
{
int N;
cin >> N;
for(int i = 1; i <= N; i )
{
long long A,B,C;
cin >> A >> B >> C;
printf("Case #%d: %sn",i, A B>C? "true" : "false");
}
return 0;
}AC代码：
代码语言：javascript复制#include <bits/stdc .h>
using namespace std;
int main()
{
int N;
cin >> N;
for(int i = 1; i <= N; i )
{
long long A,B,C;
cin >> A >> B >> C;
long long sum = A B;
if(A > 0 && B > 0 && sum < 0) //sum溢出范围后为负数
{
printf("Case #%d: truen",i);
}
else if(A < 0 && B < 0 && sum >= 0) //sum溢出范围后为正数
{
printf("Case #%d: falsen",i);
}
else
{
printf("Case #%d: %sn",i, sum>C? "true" : "false");
}
}
return 0;
}
