PAT 甲级 2020年春

7-1 Prime Day (20分)

#include<cstdio>
#include<iostream>
#include<cmath>
#include<string>
using namespace std;
bool ifprime(int x) {
	if (x < 2)return false;
        if (x == 2)return true;//2是质数
	if (x % 2 == 0)return false;
	bool ans = true;
	for (int i = 3;i <= sqrt(x);i = i + 2) {
		if (x % i == 0) {
			ans = false;
			break;
		}
	}
	return ans;
}
int main() {
	int n, i, flag = true;
	string str;
	cin >> str;
	for (i = 0;i < 8;i++) {
		n = stoi(str);
		bool ans = ifprime(n);
		if (ans) {
			cout << str << " Yes" << endl;
		}
		else {
			cout << str << " No" << endl;
			flag = false;
		}
		str.erase(str.begin());
	}
	if (flag)printf("All Prime!");
	return 0;
}

7-2 The Judger (25分)

1.sum和
2.difference差
3.product积
4.quotient商
补充:
1.surface area 表面积,volume 体积,the length of the side 边长,lateral area 侧面积,altitude高
2.rectangle长方形,square正方形,circular cylinder圆柱体dao,cone圆锥, triangle三角形,polygon多边形,rectangular solid长方体,cube立方体,circle圆
3.add,plus加 ,subtract减 ,multiply,times乘 ,divide除
参考代码

#include<iostream>
#include<set>
#include<vector>
using namespace std;
int num[20][1005];
int f[1000005] = { 0 }; //如果数字i存在,则f[i]为1。
int fout[20] = {0}, last;
vector<int>seq;//存放存在的数
int main() {
	int a, b,n,m;
	scanf("%d%d%d%d", &a, &b, &n, &m);
	last = n;
	for (int i = 0; i < n; i++)
		for (int j = 0; j < m; j++)
			scanf("%d", &num[i][j]);
	seq.push_back(a); seq.push_back(b);
	f[a] = f[b] = 1;
 
	for(int j=0;j<m;j++)
		for (int i = 0; i < n; i++) 
			if(!fout[i]){
				if (f[num[i][j]]) {
					fout[i] = 1;
					printf("Round #%d: %d is out.
", j + 1, i + 1);
					last--;
					continue;
				}
				bool flag = true;
				for(int k=0;k<seq.size();k++)
					if (seq[k] - num[i][j]>0 && f[seq[k] - num[i][j]]) { 
                                              //num[i][j]等于存在的某两数相减,相当于某数与num[i][j]相减存在。
                                              flag = false; 
                                              break; 
                                        }
				if (flag) {
					fout[i] = 1;
					printf("Round #%d: %d is out.
", j + 1, i + 1);
					last--;
					continue;
				}
				seq.push_back(num[i][j]);
				f[num[i][j]] = 1;
		}
	if (last > 0) {
		printf("Winner(s):");
		for (int i = 0; i < n; i++)if (!fout[i])printf(" %d", i + 1);
	}
	else printf("No winner."); 
	return 0;
}

自己的代码(较繁琐)

#define _CRT_SECURE_NO_WARNINGS
#include<cstdio>
#include<iostream>
#include<vector>
#include<math.h>
using namespace std;
#define MAX 1005
int or1, or2, n, m;
int G[12][MAX];
int ifout[12] = { false };
int difference[100002] = { 0 };//1为差值存在,0为不存在
bool panduan(int x, int u, int v) {
	if (difference[x] <= 0)return false;
	if (x == or1 || x == or2)return false;
	for (int j = 0; j < v; j++) {
		for (int i = 0; i < n; i++) {
			if (x == G[i][j])return false;
		}
	}
	for (int i = 0; i < u; i++)
		if (x == G[i][v])return false;
	return true;
}
void updatedif(int x, int u, int v) {
	difference[abs(or1 - x)] = 1;
	difference[abs(or2 - x)] = 1;
	for (int j = 0; j < v; j++) {
		for (int i = 0; i < n; i++) {
			if (G[i][j] > 0) difference[abs(G[i][j] - x)]++;
		}
	}
	for (int i = 0; i < u; i++)
		if (G[i][v] > 0) difference[abs(G[i][v] - x)]++;
}
int main() {
	int i, j;
	scanf("%d%d", &or1, &or2);
	difference[abs(or1 - or2)]++;
	scanf("%d%d", &n, &m);
	for (i = 0;i < n;i++) {
		for (j = 0;j < m;j++)
			scanf("%d", &G[i][j]);
	}
	for (j = 0;j < m;j++) {
		for (i = 0;i < n;i++) {
			if (ifout[i]) continue;
			if (panduan(G[i][j], i, j) == false) {//出局
				printf("Round #%d: %d is out.
", j + 1, i + 1);
				ifout[i] = j + 1;
				for (int k = j; k < m; k++)G[i][k] = -1;
			}
			else {//接受这一数字,更新所有差值
				updatedif(G[i][j], i, j);
			}
		}
	}
	bool flag = false;
	for (i = 0;i < n;i++)
		if (ifout[i] == false) {
			if (flag == false) {
				printf("Winner(s):");
				flag = true;
			}
			printf(" %d", i + 1);
		}
	if (flag == false)printf("No winner.");
	return 0;
}

7-3 Safari Park (25分)

直接用vector保存顶点对(即边)。对于图中的每一条边,在输入序列中查看两端是否为同一种动物。
用set保存输入物种序列,达到去重计数的目的。

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<set>
#include<vector>
using namespace std;
int n, m, k, r;
struct node {
	int a, b;
};
vector<node>reign;//保存图中的边
int main() {
	scanf("%d%d%d", &n, &r, &k);
	for (int i = 0; i < r; i++) {
		int a, b;
		scanf("%d%d", &a, &b);
		reign.push_back({ a,b });
	}
	scanf("%d", &m);
	for (int i = 0; i < m; i++) {
		int num[1000];//区域i的物种编号是num[i]
		set<int>iset;
		for (int j = 1; j <= n; j++) {
			scanf("%d", &num[j]);
			iset.insert(num[j]);
		}
		if (iset.size() < k) {
			printf("Error: Too few species.
"); continue;
		}
		else if (iset.size() > k) {
			printf("Error: Too many species.
"); continue;
		}
		bool flag = true;
		for (int j = 0; j < reign.size(); j++) {
			//对于图中的每一条边,在输入序列中查看两端是否为同一种动物
			int posa = reign[j].a;
			int posb = reign[j].b;
			if (num[posa] == num[posb]) { flag = false; break; }
		}
		if (!flag)printf("No
");
		else if (flag && iset.size() == k)printf("Yes
");
	}
	return 0;
}

7-4 Replacement Selection (30分)

优先队列。用结构体保存待排序的数,其中包含值(v)和它处在第几run(level)。一开始level为1,如果最近出列的数比要插入的数大,则插入的数为出列的数level+1。优先级定义为如果level小的优先级高;level一样时,值小的优先级高。

#define _CRT_SECURE_NO_WARNINGS
#include<cstdio>
#include<iostream>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
#define MAX 100005
int n, volume;
struct num {
	int v, level;
};
struct cmp {
	bool operator () (num a, num b) { 
		if (a.level != b.level)return a.level > b.level;
		else return a.v > b.v; 
	}
};
int main() {
	int i, j, nowrun = 0;
	num input, out;
	priority_queue<num, vector<num>, cmp>run;
	scanf("%d%d", &n, &volume);
	out.v = 10000000; out.level = 1;
	for (i = 0; i < n; i++) {
		if (run.size() == volume) {//队列已满,则队首出列
			out = run.top();
			if (nowrun == 0)nowrun++;
			else if (nowrun != out.level) {
                                //这一run输完了,换行输下一run,nowrun+1
				printf("
");
				nowrun++;
			}
			else printf(" ");
			printf("%d", out.v);
			run.pop();
		}
		scanf("%d", &input.v);
		if (out.v <= input.v)input.level = nowrun;
		else input.level = nowrun + 1;
		run.push(input);
	}
	while (run.size()) {//排空队列,即队列里剩下的也要全部输出
		out = run.top();
		if (nowrun == 0)nowrun++;
		else if (nowrun != out.level) {
			printf("
");
			nowrun++;
		}
		else printf(" ");
		printf("%d", out.v);
		run.pop();
	}
	return 0;
}
原文地址:https://www.cnblogs.com/yue36/p/13591070.html