#include <iostream>usingnamespacestd;
int main() {
int n;
cin >> n;
int i = 0;
int b[3] = {0};
while (n) {
b[i++] = n % 10;
n = n / 10;
}
for (int j = 0; j < b[2]; ++j) {
cout << "B";
}
for (int k = 0; k < b[1]; ++k) {
cout << "S";
}
for (int l = 1; l <= b[0]; ++l) {
cout << l;
}
return0;
}
#include <iostream>usingnamespacestd;
bool isPrime(int a) {
for (int i = 2; i * i <= a; ++i) {
if (a % i == 0)
returnfalse;
}
returntrue;
}
int main() {
int n;
cin >> n;
int count = 0;
for (int i = 5; i <= n; ++i) {
if(isPrime(i - 2) && isPrime(i))
count++;
}
cout << count;
return0;
}
#include <iostream>#include <vector>#include <algorithm>usingnamespacestd;
int main() {
int n, m;
cin >> n >> m;
vector<int> a(n);
for (int i = 0; i < n; ++i) {
cin >> a[i];
}
if (m != n && m != 0) {
if (m > n)
m = m - n;
reverse(a.begin(), a.begin() + n);
reverse(a.begin(), a.begin() + m);
reverse(a.begin() + m, a.begin() + n);
}
for (int j = 0; j < n -1; ++j) {
cout << a[j] << " ";
}
cout << a[n - 1];
return0;
}
1009. 说反话 (20)-PAT乙级真题
题目描述:
给定一句英语,要求你编写程序,将句中所有单词的顺序颠倒输出。
输入格式:测试输入包含一个测试用例,在一行内给出总长度不超过80的字符串。字符串由若干单词和若干空格组成,其中单词是由英文字母(大小写有区分)组成的字符串,单词之间用1个空格分开,输入保证句子末尾没有多余的空格。
输出格式:每个测试用例的输出占一行,输出倒序后的句子。
输入样例:
Hello World Here I Come
输出样例:
Come I Here World Hello
#include <iostream>usingnamespacestd;
int main() {
int a, b;
bool isFirst = false;
while (cin >> a >> b) {
if (b == 0)
continue;
if (isFirst)
cout << " ";
else
isFirst = true;
cout << a * b << " " << b - 1;
}
if (!isFirst)
cout << "0 0";
return0;
}
1011. A+B和C (15)-PAT乙级真题
题目描述:
给定区间[-2^31, 2^31]内的3个整数A、B和C,请判断A+B是否大于C。
输入格式:
输入第1行给出正整数T(<=10),是测试用例的个数。随后给出T组测试用例,每组占一行,顺序给出A、B和C。整数间以空格分隔。
输出格式:
对每组测试用例,在一行中输出“Case #X: true”如果A+B>C,否则输出“Case #X: false”,其中X是测试用例的编号(从1开始)。
输入样例:
4
1 2 3
2 3 4
2147483647 0 2147483646
0 -2147483648 -2147483647
输出样例:
Case #1: false
Case #2: true
Case #3: true
Case #4: false
#include <iostream>usingnamespacestd;
int main() {
int n;
cin >> n;
for (int i = 0; i < n; ++i) {
longlongint a, b, c;
cin >> a >> b >> c;
cout << "Case #" << i + 1 << ": ";
if (a + b > c)
cout << "true";
elsecout << "false";
cout << endl;
}
return0;
}