PAT甲级1065，1067题解

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本文探讨了处理64位整数溢出的策略，通过判断A+B是否大于C来解决数学问题，并提出了一种创新的排序算法，仅允许使用Swap(0,*)操作对数组进行排序，提供了详细的解题思路和代码实现。

1065 A+B and C (64bit) （20 point(s)）

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

题目大意：给三个数abc，如果a+b>c就输出yes，当然没那么简单，因为A+B的结果即使用long double都会溢出。

解题思路：两种思路吧，第一种考试实现不现实，第二种简单。第一种就是模拟大数相加了，没什么意思的就是手摸，考试时候写写很浪费时间。第二种就比较简单了，利用题目的特点。abc都是在63位范围内的，如果a+b溢出了，根据linux的特性，溢出的值返回负数，那么很简单了两个正数相加肯定比c大了，如果两个负数，变正了，那同理比c小了，如果没有溢出，那不用说了。

代码如下：

#include<iostream>
#include<string.h>
#include<vector>
#include<algorithm>
#include<iomanip>
#include<time.h>
#include<math.h>
#include<set>
#include<list>
#include<climits>
#include<queue>
#include<cstring>
#include<map>
#include<stack>
#include<string>
using namespace std;

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

	}
	return 0;
}

1067 Sort with Swap(0, i) （25 point(s)）

Given any permutation of the numbers {0, 1, 2,..., N−1}, it is easy to sort them in increasing order. But what if Swap(0, *) is the ONLY operation that is allowed to use? For example, to sort {4, 0, 2, 1, 3} we may apply the swap operations in the following way:

Swap(0, 1) => {4, 1, 2, 0, 3}
Swap(0, 3) => {4, 1, 2, 3, 0}
Swap(0, 4) => {0, 1, 2, 3, 4}

Now you are asked to find the minimum number of swaps need to sort the given permutation of the first N nonnegative integers.

Input Specification:

Each input file contains one test case, which gives a positive N (≤105) followed by a permutation sequence of {0, 1, ..., N−1}. All the numbers in a line are separated by a space.

Output Specification:

For each case, simply print in a line the minimum number of swaps need to sort the given permutation.

Sample Input:

10
3 5 7 2 6 4 9 0 8 1

Sample Output:

题目大意：一开始没看懂，后来发现是描述了一个交换规则，一个从0.。。N-1的乱序数组，交换的规则就是每次把0和值等于它的下标的值进行交换，直到交换到0的下标是0.

解题思路：一种是纯数组吧，绕一点，下面是代码。还一种应该搞两个map双向的记录然后写个while，记录一下次数就行了。

#include<iostream>
#include<string.h>
#include<vector>
#include<algorithm>
#include<iomanip>
#include<time.h>
#include<math.h>
#include<set>
#include<list>
#include<climits>
#include<queue>
#include<cstring>
#include<map>
#include<stack>
#include<string>
using namespace std;
int a[100005];
void swap(int &a, int &b) {
	int tmp = a;
	a = b;
	b = tmp;
}
int main() {
	int n;
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
	}
	int res = 0;
	for (int i = 0; i < n; i++) {
		if (a[i] != i) {
			swap(a[0], a[i]);
			if (i != 0)
				res++;
			while (a[0] != 0) {
				swap(a[0], a[a[0]]);
				res++;
			}
		}
	}
	printf("%d\n", res);
	return 0;
}