ACJoy C Beautiful Year
2016-02-27 21:01
435 查看
A B C D E F G H I J
C - Beautiful Year
Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d
& %I64u
Submit Status
Description
It seems like the year of 2013 came only yesterday. Do you know a curious fact? The year of 2013 is the first year after the old 1987 with only distinct digits.
Now you are suggested to solve the following problem: given a year number, find the minimum year number which is strictly larger than the given one and has only distinct digits.
Input
The single line contains integer y(1000 ≤ y ≤ 9000) — the year number.
Output
Print a single integer — the minimum year number that is strictly larger than y and all it's digits are distinct. It is guaranteed that the answer exists.
Sample Input
Input
Output
Input
Output
比较常规的想法,一个个往上加,直到每位数不同退出。
代码:
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
int main()
{
int year;
while(scanf("%d", &year) != EOF)
{
int res = year + 1;
int ok = 1;
int a[10];
memset(a, 0, sizeof(a));
while(ok)
{
a[0] = res/1000;
a[1] = res/100%10;
a[2] = res/10%10;
a[3] = res%10;
if(!(a[0] == a[1] || a[0] == a[2] || a[0] == a[3] || a[1] == a[2] || a[1] == a[3] || a[2] == a[3]))
{
ok = 0;
break;
}
res++;
}
printf("%d\n", res);
}
}
C - Beautiful Year
Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d
& %I64u
Submit Status
Description
It seems like the year of 2013 came only yesterday. Do you know a curious fact? The year of 2013 is the first year after the old 1987 with only distinct digits.
Now you are suggested to solve the following problem: given a year number, find the minimum year number which is strictly larger than the given one and has only distinct digits.
Input
The single line contains integer y(1000 ≤ y ≤ 9000) — the year number.
Output
Print a single integer — the minimum year number that is strictly larger than y and all it's digits are distinct. It is guaranteed that the answer exists.
Sample Input
Input
1987
Output
2013
Input
2013
Output
2014
比较常规的想法,一个个往上加,直到每位数不同退出。
代码:
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
int main()
{
int year;
while(scanf("%d", &year) != EOF)
{
int res = year + 1;
int ok = 1;
int a[10];
memset(a, 0, sizeof(a));
while(ok)
{
a[0] = res/1000;
a[1] = res/100%10;
a[2] = res/10%10;
a[3] = res%10;
if(!(a[0] == a[1] || a[0] == a[2] || a[0] == a[3] || a[1] == a[2] || a[1] == a[3] || a[2] == a[3]))
{
ok = 0;
break;
}
res++;
}
printf("%d\n", res);
}
}
相关文章推荐
- 使用C++实现JNI接口需要注意的事项
- 关于指针的一些事情
- c++ primer 第五版 笔记前言
- share_ptr的几个注意点
- Lua中调用C++函数示例
- Lua教程(一):在C++中嵌入Lua脚本
- Lua教程(二):C++和Lua相互传递数据示例
- C++联合体转换成C#结构的实现方法
- C++高级程序员成长之路
- C++编写简单的打靶游戏
- C++ 自定义控件的移植问题
- C++变位词问题分析
- C/C++数据对齐详细解析
- C++基于栈实现铁轨问题
- C++中引用的使用总结
- 使用Lua来扩展C++程序的方法
- C++中调用Lua函数实例
- Lua和C++的通信流程代码实例
- C与C++之间相互调用实例方法讲解
- 解析C++中派生的概念以及派生类成员的访问属性