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 = 1; //变量不能用case scanf("%d",&k); for(int i = 0; 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 < 0 && b < 0 && res >= 0) flag = false; // 负溢出时等号可以取到 else if(a > 0 && b > 0 && res < 0) 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 0; }