HUST 1602 - Substring

ShenduCC

发布于 2018-04-26 11:52:44

7680

发布于 2018-04-26 11:52:44

文章被收录于专栏：算法修养

题目描述 This problem is quiet easy. Initially, there is a string A.

Then we do the following process infinity times. A := A + “HUSTACM” + A

For example, if a = “X”, then After 1 step, A will become “XHUSTACMX” After 2 steps, A will become “XHUSTACMXHUSTACMXHUSTACMX”

Let A = “X”, Now I want to know the characters from L to R of the final string. 输入 Multiple test cases, in each test case, there are only one line containing two numbers L and R. 1 <= L <= R <= 10^12 R-L <= 100 输出 For each test case, you should output exactly one line which containing the substring. 样例输入 5 10 样例输出 TACMXH

#include <iostream>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <stdlib.h>
#include <stdio.h>

using namespace std;
long long int l,r;
char a[9];
int main()
{
    a[0]='H';
    a[1]='U';
    a[2]='S';
    a[3]='T';
    a[4]='A';
    a[5]='C';
    a[6]='M';
    a[7]='X';
    while(scanf("%lld%lld",&l,&r)!=EOF)
    {
        int pos;
        if(l==1)
            pos=7;
        else
            pos=(l-2)%8;
         int cnt=1;
         while(cnt<=(r-l)+1)
         {
             printf("%c",a[(pos)%8]);
             pos++;
             cnt++;
         }
         printf("\n");


    }
    return 0;
}

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原始发表：2016-03-09 ，如有侵权请联系 cloudcommunity@tencent.com 删除

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