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
时间限制
100 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
HOU, Qiming

知识点：

• 正溢出：两个正数和等于负数，负溢出：两个负数和等于正数
• long long 范围：[-2^63,2^63)
• A+B最大是2^64-2，溢出后为-2,因为(2^64-2)%(2^64)
• 所以A>0,B>0,A+B<0时是正溢出，此时A+B>C
• A<0,B<0,A+B>=0,负溢出，A+B

#include<cstdio>int main(){    int n,i;    long long a,b,c,d;    scanf("%d",&n);    for(i=0;i<n;i++)    {    scanf("%lld%lld%lld",&a,&b,&c);          d=a+b;    if(a>0&&b>0&&d<=0) printf("Case #%d: true\n",i+1);    else if(a<0&&b<0&&d>=0) printf("Case #%d: false\n",i+1);    else {if(d>c)        printf("Case #%d: true\n",i+1);    else printf("Case #%d: false\n",i+1);}    }    return 0;}