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

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

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:

31 2 32 3 49223372036854775807 -9223372036854775808 0

Sample Output:

Case #1: falseCase #2: trueCase #3: false

AC代码1 long double 不用判断溢出

#include<iostream>#include<string>#include<cstdio>using namespace std;bool judge(string ,string ,long long);int main() {    int n,count=1;    cin>>n;    long double a,b,c;    while(n--) {        cin>>a>>b>>c;        bool ans = a+b >c;        printf("Case #%d: ",count++);        printf(ans?"true\n":"false\n");    }    return 0;}

AC代码2 longlong 判断溢出

#include<iostream>#include<string>#include<cstdio>using namespace std;bool judge(string ,string ,long long);int main() {    int n,count=1;    cin>>n;    long long a,b,c,sum;    while(n--) {        cin>>a>>b>>c;        bool ans;        sum = a+b;        if(a<0&&b<0&&sum>=0)//负溢出            ans =false;        else if(a>0&&b>0&&sum<=0)            ans = true;//正溢出        else            ans = sum>c;        printf("Case #%d: ",count++);        printf(ans?"true\n":"false\n");    }    return 0;}