codeforces_664C. International Olympiad

最新推荐文章于 2017-05-24 15:26:15 发布

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最新推荐文章于 2017-05-24 15:26:15 发布

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本文介绍了一种算法，用于确定国际奥林匹克竞赛的年份，该算法基于竞赛名称中的数字部分进行计算。从1989年开始，每年的竞赛名称都会包含一个唯一的数字序列，代表该年的最后几位数字。通过理解数字序列的规律，可以准确地推算出对应年份。

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International Abbreviation Olympiad takes place annually starting from 1989. Each year the competition receives an abbreviation of formIAO'y, where y stands for some number of consequent last digits of the current year. Organizers always pick an abbreviation with non-empty string y that has never been used before. Among all such valid abbreviations they choose the shortest one and announce it to be the abbreviation of this year's competition.

For example, the first three Olympiads (years 1989, 1990 and 1991, respectively) received the abbreviations IAO'9, IAO'0 andIAO'1, while the competition in 2015 received an abbreviation IAO'15, as IAO'5 has been already used in 1995.

You are given a list of abbreviations. For each of them determine the year it stands for.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 1000) — the number of abbreviations to process.

Then n lines follow, each containing a single abbreviation. It's guaranteed that each abbreviation contains at most nine digits.

Output

For each abbreviation given in the input, find the year of the corresponding Olympiad.

Examples

input
5
IAO'15
IAO'2015
IAO'1
IAO'9
IAO'0

output
2015
12015
1991
1989
1990

input
4
IAO'9
IAO'99
IAO'999
IAO'9999

output
1989
1999
2999
9999
这道题wa了很多次，后来找到规律后才过了，长度为n的数字字符串的总个数为10^(n-1)个，
通过这个规律算出当前数字字符串对应的年数是从1989开始的第几年，之后加上1989就可以了。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <stack>
#include <bitset>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <algorithm>
#define Si(a) scanf("%d",&a)
#define Sl(a) scanf("%lld",&a)
#define Sd(a) scanf("%lf",&a)
#define Ss(a) scanf("%s",a)
#define Pi(a) printf("%d\n",(a))
#define Pl(a) printf("%lld\n",(a))
#define Pd(a) printf("%lf\n",(a))
#define Ps(a) printf("%s\n",(a))
#define W(a) while(a--)
#define mem(a,b) memset(a,(b),sizeof(a))
#define FOP freopen("data.txt","r",stdin)
#define inf 0x3f3f3f3f
#define maxn 1000010
#define mod 1000000007
#define PI acos(-1.0)
#define LL long long
using namespace std;

LL dp[20];

LL get_m(int n)
{
    LL ans=1;
    while(n--)ans*=10;
    return ans;
}

LL get_num(int n)
{
    if(n==1)return 9;
    else if(n==2||n==3)return 99;
    else if(n==4)return 3099;
    else if(n>4)
    {
        n-=5;
        LL num=10000;
        LL ans=num;
        while(n--)
        {
            ans+=num*10;
            num*=10;
        }
        return ans+3099;
    }
}

int main()
{
    int i;
    mem(dp,0);
    for(i=2;i<10;i++)
    {
        dp[i]=dp[i-1]+get_m(i-1);
    }
    //FOP;
    int T;
    Si(T);
    while(T--)
    {
        char a[30];
        LL ans=0;
        Ss(a);
        int len=strlen(a)-4;
        LL n=0;
        for(i=4;a[i]!='\0';i++)
        {
            n=n*10+a[i]-'0';
        }
        LL numl=get_num(len);
        if(n>=numl)
        {
            ans=1989+dp[len]+(n-numl);
        }
        else
        {
            ans=1989+dp[len]+(get_m(len)+n-numl);
        }
        Pl(ans);
    }
}