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.\n", 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.\n", 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.\n", 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.\n"); continue;
}
else if (iset.size() > k) {
printf("Error: Too many species.\n"); 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\n");
else if (flag && iset.size() == k)printf("Yes\n");
}
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("\n");
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("\n");
nowrun++;
}
else printf(" ");
printf("%d", out.v);
run.pop();
}
return 0;
}
