PAT(A) 1065. A+B and C (64bit) (20)

Given three integers A, B and C in [-2⁶³, 2⁶³], 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:

3
1 2 3
2 3 4
9223372036854775807 -9223372036854775808 0

Sample Output:

Case #1: false
Case #2: true
Case #3: false

using namespace std;
#include <cstdio>

//about 大整数溢出 注意long long的范围：[-2^63, 2^63) 所以可能溢出
int main()
{
    int T, num=1;
    scanf("%d", &T);
    while(T--)
    {
        long long a, b, c;     //long long 范围：[-2^53, 2^63)
        scanf("%lld%lld%lld", &a, &b, &c);     //注(1)：long long 对应 %lld
        long long sum=a+b;     //注(2)：a+b必须赋值给一个long long 变量再比较才能正确输出
        bool flag;     　　　　 //开始判断各种情况:

        //划重点(1)：当a+b>=2^63时，显然 a+b>c成立，但a+b会超过long long的正向最大值而发生正溢出
        //由范围=> a+b的最大值为2^64-2, 所以 long long存储正溢出后的值的区间为 [-2^63, -2]
　　　　 //由(2^64-2)%2^64=-2 =>右边界
        if(a>0 && b>0 && sum<0)         flag=true;   //正溢出为true

        //划重点(2)：当a+b<2^63时，显然 a+b<c成立，但a+b会超过long long的负向最大值而发生负溢出
        //由范围=> a+b的最小值为-2^64, 所以 long long存储负溢出后的值的区间为 [0, 2^63)
　　　　 //由(-2^64)%2^64=0 =>左边界
        else if(a<0 && b<0 && sum>=0)   flag=false;  //负溢出为false

        else if(sum>c)  flag=true;  //无溢出时，A+B>C时为true
        else flag=false;

        if(flag==true)
            printf("Case #%d: true\n", num++);
        else
            printf("Case #%d: false\n", num++);
    }
    return 0;
}