Codeforces Round #419 (Div. 2) A. Karen and Morning(模拟)
http://codeforces.com/contest/816/problem/A
题意:
给出一个时间,问最少过多少时间后是回文串。
思路:
模拟,先把小时的逆串计算出来:
① 如果逆串=分钟,那么此时已经是回文串了。
② 如果逆串>分钟,那么只需要逆串-分钟即可。(注意此时逆串>=60的情况)
③ 如果逆串<分钟,此时在这个小时内要构成回文串已经是不可能的了,那么就加上60-minute分钟,进入一个新的小时,然后重复上述步骤。
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<sstream>
#include<vector>
#include<stack>
#include<queue>
#include<cmath>
#include<map>
#include<set>
using namespace std;
typedef long long ll;
typedef pair<int,int> pll;
const int INF = 0x3f3f3f3f;
const int maxn=*1e5+; char str[]; int main()
{
//freopen("in.txt","r",stdin);
while(~scanf("%s",&str))
{
int hour = (str[]-'')* + (str[]-'');
int minute = (str[]-'')*+(str[]-'');
int rev_h = (str[]-'')*+str[]-'';
int rev_m = (str[]-'')*+(str[]-''); if(hour==rev_m) {puts("");continue;} int ans=;
while(true)
{
rev_h = (hour%*) + hour/;
if(rev_h == minute) break;
if(rev_h>minute && rev_h<)
{
ans+=rev_h-minute;
break;
} else
{
ans+=-minute;
minute=;
hour=(hour+)%;
}
}
printf("%d\n",ans);
}
return ;
}
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