春節刷題day1[1010 一元多項式求導 ]

阿新 • • 發佈：2021-02-07

技術標籤：刷題指南 c++

春節刷題day1：PAT

1001 害死人不償命的(3n+1)猜想

1002 寫出這個數

1003 我要通過！

1004 成績排名

1005 繼續(3n+1)猜想

1006 換個格式輸出整數

1007 素數對猜想

1008 陣列元素迴圈右移問題

1009 說反話

1010 一元多項式求導

1、1001 害死人不償命的(3n+1)猜想

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<map> 

#include<stack>
#include<queue>
#include<vector>
#include<string>
#include<cstring>

using namespace std;

int n, m, T;

int main(){
	int i, j, k;
	while(cin >> n){
		int ans = 0;
		while(n != 1){
			if(n & 1) n = (3 * n + 1) / 2;
			else n /= 2;
			ans++;
		}
		cout << 
 ans << endl;
	}
	return 0;
}

2、1002 寫出這個數

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<string>
#include<cstring> 


using namespace std;

int n, m, T, tot;
string s, ans[15];
map<int, string> p;

void init(){
	p[0] = "ling"; p[1] = "yi"; p[2] = "er"; p[3] = "san";
	p[4] = "si"; p[5] = "wu"; p[6] = "liu"; p[7] = "qi";
	p[8] = "ba"; p[9] = "jiu";
}

int main(){
	int i, j, k;
	init();
	while(cin >> s){
		int len = s.size(), sum = 0;
		for(i = 0; i < len; i++) sum += s[i] - '0';
		while(sum){
			ans[tot++] = p[sum % 10];
			sum /= 10;
		}
		reverse(ans, ans + tot);
		for(i = 0; i < tot; i++){
			cout << ans[i];
			printf("%c", i == tot - 1 ? '\n' : ' ');
		}
	}
	return 0;
}

3、1003 我要通過！

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<string>
#include<cstring>

using namespace std;

int n, m, T, len;
string s;

bool checka(){
	for(int i = 0; i < len; i++){
		if(s[i] != 'P' && s[i] != 'A' && s[i] != 'T') return false;
	}
	return true;
}

bool checkb(){
	int l, r, num = 0;
	for(int i = 0; i < len; i++)
		if(s[i] == 'P'){ l = i; break; }
	for(int i = len - 1; i >= 0; i--)
		if(s[i] == 'T'){ r = i; break; }
	for(int i = 0; i < len; i++) if(s[i] == 'A') num++;
	if((num != len - 2) || (len - 2 == 0)) return false;
	if((r - l - 1) * l != len - r - 1) return false;
	return true;
}

int main(){
	int i, j, k;
	cin >> T;
	while( T-- ){
		cin >> s; len = s.size();
		if(checka() && checkb()) puts("YES");
		else puts("NO");
	}
	return 0;
}

4、1004 成績排名

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<string>
#include<cstring>

using namespace std;

int n, m, T, len;

struct node{
	string name;
	string num;
	int score;
	bool operator < (const node &A)const{
		return score < A.score;
	}
}a[10005];

int main(){
	int i, j, k;
	while( cin >> n){
		for(i = 0; i < n; i++){
			cin >> a[i].name >> a[i].num >> a[i].score;
		}
		sort(a, a + n);
		cout << a[n - 1].name << " " << a[n - 1].num << endl;
		cout << a[0].name << " " << a[0].num << endl;
	}
	return 0;
}

5、1005 繼續(3n+1)猜想

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<string>
#include<cstring>
#define fi first
#define se second

using namespace std;

int n, m, T, tot;
map<int, int> p;
map<int, int>::iterator it;
int ans[105];

int main(){
	int i, j, k;
	while( cin >> n){
		for(i = 0; i < n; i++){
			cin >> m;
			p[m] = 1;
		}
		for(it = p.begin(); it != p.end(); it++){
			k = it -> fi;
			while(k != 1){
				if(p[k]) p[k]++;
				if(k & 1) k = (3 * k + 1) / 2;
				else k /= 2;
			}
		}
		for(it = p.begin(); it != p.end(); it++){
			if(it -> se == 2) ans[tot++] = it -> fi;
		}
		sort(ans, ans + tot, greater<int>());
		for(i = 0; i < tot; i++)
			printf("%d%c", ans[i] , i == tot - 1 ? '\n' : ' ');
	}
	return 0;
}

6、1006 換個格式輸出整數

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<string>
#include<cstring>
#define fi first
#define se second

using namespace std;

int n, m, T, tot;
string s;

int main(){
	int i, j, k;
	while( cin >> n){
		for(i = 0; i < n / 100; i++) printf("B");
		for(i = 0; i < n % 100 / 10; i++) printf("S");
		for(i = 1; i <= n % 10; i++) printf("%d", i);
		puts("");
	}
	return 0;
}

7、1007 素數對猜想

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<string>
#include<cstring>
#define fi first
#define se second

using namespace std;

int n, m, T, tot;
int prime[100005];

bool check(int x){
	for(int i = 2; i <= sqrt(x); i++)
		if(x % i == 0) return false;
	return true;
}

int main(){
	int i, j, k;
	for(i = 2; i <= 100005; i++){ 
		if(check(i) ) prime[tot++] = i;
	}
	while(cin >> n){
		int ans = 0;
		for(i = 1; prime[i] <= n; i++){
			if(prime[i] - prime[i - 1] == 2) ans++;
		}
		cout << ans << endl;
	}
	return 0;
}

8、1008 陣列元素迴圈右移問題

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<string>
#include<cstring>
#define fi first
#define se second

using namespace std;

int n, m, T, tot;
vector<int> v;

int main(){
	int i, j, k;
	while(cin >> n >> m){
		for(i = 0; i < n; i++){
			cin >> k; v.push_back(k);
		}
		m %= n;
		if(m){
			printf("%d", v[n - m]);
			for(i = n - m + 1; i < n; i++) printf(" %d", v[i]);
			for(i = 0; i < n - m; i++) printf(" %d", v[i]);
		}else{
			printf("%d", v[0]);
			for(i = 1; i < n; i++) printf(" %d", v[i]);
		}
		puts("");
	}
	return 0;
}

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<string>
#include<cstring>
#define fi first
#define se second

using namespace std;

int n, m, T, tot;
int a[105];
vector<int> v;

int main(){
	int i, j, k;
	while(cin >> n >> m){
		for(i = 0; i < n; i++){
			cin >> k; v.push_back(k);
		}
		m %= n;
		reverse(v.begin() + n - m, v.end());
		reverse(v.begin(), v.begin() + n - m);
		reverse(v.begin(), v.end());
		for(i = 0; i < n; i++)
			printf("%d%c", v[i], i == n - 1 ? '\n' : ' ');
	}
	return 0;
}

9、1009 說反話

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<sstream>
#include<string>
#include<cstring>
#define fi first
#define se second

using namespace std;

int n, m, T, tot;
string s, ans[105];

int main(){
	int i, j, k;
	while( getline(cin, s)){
		string x;
		stringstream ss(s);
		while(ss >> x) ans[tot++] = x;
		for(i = tot - 1; i >= 0; i--){
			cout << ans[i];
			printf("%c", i == 0 ? '\n' : ' ');
		}
	}
	return 0;
}

10、1010 一元多項式求導

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<sstream>
#include<string>
#include<cstring>
#define fi first
#define se second

using namespace std;

int n, m, T, tot;
int a[10005], b[10005];

int main(){
	int i, j, k;
	bool ok = false;
	for(i = 0; ; i++){
		cin >> a[i] >> b[i];
		if(b[i] == 0) break;
		a[i] *= b[i]; b[i] -= 1; ok = true;
	}
	for(j = 0; j < i; j++)
		printf("%d %d%c", a[j], b[j], j == i - 1 ? '\n' : ' ');
	if(!ok) printf("0 0\n");
	return 0;
}