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

//long long 有效位为[-2^63，2^63,)，如果a，b取到临界点，会出现溢出。

// A+B最大为 2^64-2（左开右闭），（2^64-2）%2^63=-2，long long的正溢出的范围为[-2^63,-2]

// 所以A>0,B>0,A+B<0时，为正溢出，结果为true。

//相同long long负溢出范围为[0,2^63],当A<0,B<0,A+B>=0时结果为负

#include<cstdio>

int main(){

    int k,Tcase = ; //变量不能用case

    scanf("%d",&k);

    for(int i = ; i < k; i++){

        long long a,b,c;

        scanf("%lld%lld%lld",&a,&b,&c);

        long long res;

        res = a + b;

        bool flag;

        if(a <  && b <  && res >= ) flag = false;   // 负溢出时等号可以取到

        else if(a >  && b >  && res < ) flag = true;  //正溢出为true，负溢出为false

        else if(res > c) flag = true;

        else flag = false;

        if(flag == true) printf("Case #%d: true\n",Tcase++);

        else printf("Case #%d: false\n",Tcase++);

    }

    return ;

}

巴特西

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

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