PAT甲级刷题记录-(AcWing)-Day08图论(8题)
课程来源AcWing
其中AcWing中的题目为翻译好的中文题目
刷题列表
1019 General Palindromic Number
注意点
PAT上没有关于0的判断, 其他的思路不难, 先转换进制然后再判断一下输出就好
#include <iostream>
#include <vector>
using namespace std;
vector<int> res;
bool check(int n, int b) {
if (n == 0) {
res.push_back(0);
return true;
}
while (n) res.push_back(n % b), n = n / b;
for (int i = 0, j = res.size() - 1; i < j; ++i, --j) {
if (res[i] != res[j])
return false;
}
return true;
}
int main() {
int n, b;
cin >> n >> b;
if (check(n, b)) puts("Yes");
else puts("No");
cout << res[res.size() - 1];
for (int i = res.size() - 2; i >= 0; i--) {
cout << " " << res[i];
}
return 0;
}
1049 Counting Ones
英语单词
解析
具体可以看<编程之美>的2.4节内容
编程之美链接
链接：https://pan.baidu.com/s/1bIRMzUfSCyIqFAkwI4qEug?pwd=sbbi
提取码：sbbi
–来自百度网盘超级会员V3的分享
注意点
#include <iostream>
#include <vector>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> num;
int res = 0;
// 使用vector记录下数字n的每一位
while (n) num.push_back(n % 10), n = n / 10;
// 计算每一位中1的个数
for (int i = num.size()-1; i >= 0; i--) {
int d = num[i];
int high = 0, low = 0, power = 1; // high记录d左边高位的个数,low记录d右边低位的个数,power来记录要乘10的几次
for (int j = num.size() - 1; j > i; j--) {
// 算高位high有几个
high = high * 10 + num[j];
}
for (int j = i - 1; j >= 0; j--) {
// 算低位有几个,并记录位数(10的几次)
low = low * 10 + num[j];
power *= 10;
}
if (d == 0) res += high * power;
else if (d == 1) res += high * power + low + 1;
else res += (high + 1) * power;
}
cout << res << endl;
return 0;
}
#include <iostream>
#include <vector>
using namespace std;
int cal(int n) {
int res = 0;
vector<int> num;
for (auto &c:to_string(n)) {
num.push_back(c - '0');
}
for (int i = 0; i < num.size(); i++) {
int d = num[i];
int left = 0, right = 0, power = 1;
for (int j = 0; j < i; j++) {
left = left * 10 + num[j];
}
for (int j = i + 1; j <num.size(); ++j) {
right = right * 10 + num[j];
power *= 10;
}
if (d == 0) res += left * power;
else if (d == 1) res += left * power + right + 1;
else res += (left + 1) * power;
}
return res;
}
int main() {
int n;
cin >> n;
cout << cal(n) << endl;
return 0;
}
1059 Prime Factors
注意点
要注意最后结束循环的时候剩下的那个数也是质数,要一起输出
#include <iostream>
using namespace std;
int main() {
int n;
cin >> n;
cout << n << "=";
bool is_first = true;
for (int i = 2; i <= n / i; ++i) {
if (n % i == 0) {
int k = 0;
while (n % i == 0) n = n / i, k++;
if (is_first) is_first = false;
else cout << "*";
if (k > 1)
cout << i << "^" << k;
else
cout << i;
}
}
if (is_first) cout << n << endl;
else if (n != 1) cout << "*" << n << endl;
return 0;
}
1081 Rational Sum
英语单词
- numerators and denominators 分子和分母
- rational numbers 有理数
注意点
用long long来存储,然后在运算的过程中多求约分
#include <iostream>
using namespace std;
typedef long long LL;
const int N = 110;
LL gcd(LL a, LL b) {
return b ? gcd(b, a % b) : a;
}
int main() {
int n;
scanf("%d", &n);
LL a = 0, b = 1, t;
for (int i = 0; i < n; ++i) {
LL c, d;
scanf("%lld/%lld", &c, &d);
t = gcd(b, d);
a = d / t * a + b / t * c;
b = b / t * d;
t = gcd(a, b);
a /= t, b /= t;
}
if (b == 1) cout << a;
else {
if (a >= b) printf("%lld ", a / b);
printf("%lld/%lld", a % b, b);
}
return 0;
}
1096 Consecutive Factors
英语单词
解析
注意点
这里要注意到
#include <iostream>
#include <vector>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> res;
for (int i = 2; i <= n / i; ++i) {
vector<int> seq;
if (n % i == 0) {
int t = n;
int j = i;
while (t % j == 0) {
seq.push_back(j);
t = t / j++;
}
if (seq.size() > res.size()) res = seq;
}
}
if (res.empty()) {
res.push_back(n);
}
cout << res.size() << endl;
cout << res[0];
for (int i = 1; i < res.size(); ++i) {
cout << "*" << res[i];
}
return 0;
}
1103 Integer Factorization
英语单词
- factorization 因式分解
解析
注意点
这题用DP做,还没学,今天先空着
1104 Sum of Number Segments
英语单词
- consecutive 连续的
解析
找规律,找每个点会被加几次
注意点
double的精度不太够,long double就可以了
#include<iostream>
#include <cstdio>
using namespace std;
int main() {
int n;
cin >> n;
long double res = 0;
for (int i = 1; i <= n; ++i) {
long double x;
cin >> x;
res += x * i * (n - i + 1);
}
printf("%.2Lf\n", res);
return 0;
}
1088 Rational Arithmetic
注意点
#include <iostream>
#include <cstring>
typedef long long LL;
using namespace std;
LL a, b, c, d;
LL gcd(LL x, LL y) {
return y ? gcd(y, x % y) : x;
}
void print(LL res_a, LL res_b) {
if (res_b < 0) res_a *= -1, res_b *= -1;
LL t = gcd(abs(res_a), res_b);
res_a /= t, res_b /= t;
if (res_a < 0) printf("(");
if (res_b == 0) printf("Inf");
else if (res_a == 0) printf("0");
else if (res_b == 1) printf("%lld", res_a);
else if (abs(res_a) > res_b) printf("%lld %lld/%lld", res_a / res_b, abs(res_a) % res_b, res_b);
else printf("%lld/%lld", res_a, res_b);
if (res_a < 0) printf(")");
}
//void print_one(int num1,int num2) {
// if (num2 == 0) printf("Inf\n");
// else if (num1 == 0) printf("0\n");
// else if (num2 == 1) printf("%lld\n", num1);
// else if (num1 > num2) printf("%lld %lld/%lld\n", num1 / num2, num1 % num2, num2);
// else printf("%lld/%lld\n", num1, num2);
//
//}
void add() {
print(a, b);
printf(" + ");
print(c, d);
printf(" = ");
LL res_a = 0, res_b = 0;
LL t = gcd(b, d);
res_a = d / t * a + b / t * c;
res_b = b / t * d;
print(res_a, res_b);
printf("\n");
}
void sub() {
print(a, b);
printf(" - ");
print(c, d);
printf(" = ");
LL res_a = 0, res_b = 0;
LL t = gcd(b, d);
res_a = d / t * a - b / t * c;
res_b = b / t * d;
print(res_a, res_b);
printf("\n");
}
void mul() {
print(a, b);
printf(" * ");
print(c, d);
printf(" = ");
LL res_a = 0, res_b = 0;
res_a = a * c;
res_b = b * d;
print(res_a, res_b);
printf("\n");
}
void div() {
print(a, b);
printf(" / ");
print(c, d);
printf(" = ");
LL res_a = 0, res_b = 0;
res_a = a * d;
res_b = b * c;
print(res_a, res_b);
printf("\n");
}
int main() {
scanf("%lld/%lld %lld/%lld", &a, &b, &c, &d);
add();
sub();
mul();
div();
return 0;
}
模板
英语单词
解析
注意点