A1065 A+B and C (64bit)
decription
Given three integers A, B and C in [-2^63, 2^63], 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
idea
- A,B的范围为 [-2^63, 2^63]
则A+B可能会超出long long 范围
结合本题只有20分,不至于用到大数运算 - 考虑分类讨论,记sum = A+B
当sum正溢出时,sum<0,则为true
当sum负溢出时,sum>=0,则为false(注意等于的情况,否则第三个测试点过不去
当sum为发生溢出时,直接比较sum和c的大小即可
solution
#include <stdio.h>
int main(){
int n;
scanf("%d", &n);
for(int i = 1; i <= n; i++){
long long a, b, c, sum;
scanf("%lld%lld%lld", &a, &b, &c);
sum = a + b;
if(sum < 0 && a > 0 && b > 0)
printf("Case #%d: true", i);
else if (sum >= 0 && a < 0 && b < 0)
printf("Case #%d: false", i);
else{
if(sum > c)
printf("Case #%d: true", i);
else
printf("Case #%d: false", i);
}
if(i != n)
printf("\n");
}
return 0;
}