简单题. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<stack> #include<queue> #include<string> #include<algorithm> using namespace std; int n; ]; int num1,num2;…

PAT (Advanced Level) Practice 1023 Have Fun with Numbers (20 分) 凌宸1642 题目描述: Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be…

题意: 输入一个四位的正整数N,输出每位数字降序排序后的四位数字减去升序排序后的四位数字等于的四位数字,如果数字全部相同或者结果为6174(黑洞循环数字)则停止. trick: 这道题一反常态的输入的数字是一个int类型而不是包含前导零往常采用字符串的形式输入,所以在测试点2,3,4如果用字符串输入会超时..... AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace st…

题目信息 1069. The Black Hole of Numbers (20) 时间限制100 ms 内存限制65536 kB 代码长度限制16000 B For any 4-digit integer except the ones with all the digits being the same, if we sort the digits in non-increasing order first, and then in non-decreasing order, a new n…

PAT (Advanced Level) Practice 1031 Hello World for U (20 分) 凌宸1642 题目描述: Given any string of N (≥5) characters, you are asked to form the characters into the shape of U. For example, helloworld can be printed as: h d e l l r lowo That is, the charact…

1069 The Black Hole of Numbers (20 分)   For any 4-digit integer except the ones with all the digits being the same, if we sort the digits in non-increasing order first, and then in non-decreasing order, a new number can be obtained by taking the seco…

1069 The Black Hole of Numbers (20分) 1. 题目 2. 思路 把输入的数字作为字符串,调用排序算法,求最大最小 3. 注意点 输入的数字的范围是(0, 104), 如果作为字符串处理时要注意前面补0 4. 代码 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; bool cmp(char a, char b){ return a>…

题目 For any 4-digit integer except the ones with all the digits being the same, if we sort the digits in nonincreasing order first, and then in non-decreasing order, a new number can be obtained by taking the second number from the first one. Repeat i…

For any 4-digit integer except the ones with all the digits being the same, if we sort the digits in non-increasing order first, and then in non-decreasing order, a new number can be obtained by taking the second number from the first one. Repeat in…

因为会溢出,因此判断条件需要转化.变成b>c-a #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<queue> #include<algorithm> using namespace std; long long a,b,c; int main() { int T; scanf("%d",&T);…