0.题目链接https://www.patest.cn/contests/pat-a-practise/1065
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 0Sample Output:
Case #1: false Case #2: true Case #3: false
1.题意分析
给出三个整数A,B,C，范围在【-2^63,2^63】之间，判断A+B是否大于C.
2.坑点分析
（1）A,B,C三个数据都是【-2^63,2^63】之间，所以这道题需要考虑大数据，即我们必须使用long long 来存储数据。
（2）对于两个数A,B，我们可以知道只有当A和B同时大于0时，才可能有A+B>2^63;当A和B同时小于0的时候，才有A+B<(-2^63)；若不在此范围之内，我们肯定可以保证数据不会越界。
3.代码如下
#include <cstdio>
long long inf1 = ((long long)1<<63)-1;//最大值
long long inf2 = (-1)*inf1- 1;//最小值
void function(long long result ,long long a,long long b,long long c,int count){
result = a+b;
if(result <= c){
printf("Case #%d: false\n",count);
}
else
{
printf("Case #%d: true\n",count);
}
count++;
}
int main(){
int number;
scanf("%d",&number);
int count = 1;
while(number--){
long long a ,b ,c;
long long result ;
scanf("%lld %lld %lld",&a,&b,&c);
//如果两者同时小于0，则有可能超过下界
if(a<0 && b<0){
if(inf2 - a > b){
printf("Case #%d: false\n",count);//a+b肯定小于c
}
else{
function(result,a,b,c,count);
}
}
//如果两者同时大于0，则有可能超过上界
else if(a >0 && b>0){
if(inf1 - a < b){
printf("Case #%d: true\n",count);//a+b肯定大于c
}
else{
function(result,a,b,c,count);
}
}
else{
function(result,a,b,c,count);
}
count++;
}
}
/**测试数据
3
0 0 0
2 3 4
-9223372036854775807 -9223372036854775808 0
**/
