看个人思路吧,清晰的话就简单 CodeForces 271A - Beautiful Year
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 test(s)
input
1987
output
2013
input
2013
output
2014
题意就是说给你一个年份,当然是四位数的,然后要你找出比这个数大且是最先打到四位数每一个数字都不一样,可能有些同志英语不好看不懂然后小弟语文差表述也有问题,这样说吧:2013是给定的数字,比2013大且最先达到每一个数字都不一样就是2014 2 0 1 4;如今应该能明确了吧?
那咱们如今就来讲讲思路:
定义一个year作为输入,然后就是while(1)的一个循环,再定义一个flag作为标记,初值为0;
然后就是取各个数位上的数字分别赋值给a,b,c,d,然后就是一个if比較,假设不一样则flag=1;
然后推断flag是否为一,为一代表已找到符合条件的数字,则break;
这里有一点注意的是:输入year之后一定要加一,不然像2013这种是会输出2013滴!!!!!!切记,好啦,讲完!!!!!!
贴下代码:
#include<stdio.h> #include<string.h> #include<iostream> using namespace std; int main() { int i,j,k; int t,n,m; int a,b,c,d; int year,flag; while(scanf("%d",&year)!=EOF) { flag=0; //printf("%d %d %d %d\n",a,b,c,d); while(1) { year++; //千万注意这里!!!!!!千万千万千千万!!!!!! d=year%10; c=(year/10)%10; b=(year/100)%10; a=(year/1000); if(a!=b&&a!=c&&a!=d&&b!=c&&b!=d&&c!=d) { flag=1; } if(flag==1) break; } printf("%d\n",year); } return 0; }
如有发现bug,就指出来吧!!!!!!不胜感激!!!!!!