PAT (Advanced Level) Practise 1065 A+B and C (64bit) (20)

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:

3
1 2 3
2 3 4
9223372036854775807 -9223372036854775808 0

Sample Output:

Case #1: false
Case #2: true
Case #3: false

故意卡在longlong的上限上，可以感受到出题人满满的恶意，先读在字符串里，然后去掉符号用unsignedlonglong来存，剩下的就是判断符号和越界的问题。
#include<cmath>
#include<queue>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
const int mod = 1e9 + 7;
const int maxn = 1e3 + 10;
const ull INF = ((ull)1 << 63);
int T, t;
char s[maxn];
ull a, b, c;

bool check()
{
	int f1 = 0, f2 = 0, f3 = 0;
	scanf("%s", s);	f1 = s[0] == '-'; sscanf(f1 ? s + 1 : s, "%llu", &a);
	scanf("%s", s);	f2 = s[0] == '-'; sscanf(f2 ? s + 1 : s, "%llu", &b);
	scanf("%s", s);	f3 = s[0] == '-'; sscanf(f3 ? s + 1 : s, "%llu", &c);
	if (a == INF && a == b && f1 == f2) return !f1;
	if (f1 == f2) b += a; 
	else
	{
		if (f1) if (a > b) { f2 = 1; b = a - b; } else b -= a;
		else if (a >= b) { f2 = 0; b = a - b; } else b -= a;
	}
	return f2 == f3 ? f2 ? b < c : b > c : f2 < f3;
}

int main()
{
	scanf("%d", &T);
	while (T--)
	{
		printf("Case #%d: %s\n", ++t, check() ? "true" : "false");
	}
	return 0;
}