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hust 1602, 乱搞

2013-11-07 18:41 423 查看
D - Substring
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu
Submit Status Practice HUST 1602

Description

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.

Input

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

Output

For each test case, you should output exactly one line which containing the substring.

Sample Input

5 10


Sample Output

TACMXH


#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<vector>
#include<cstdlib>
#include<algorithm>

using namespace std;

#define LL long long
#define ULL unsigned long long
#define UINT unsigned int
#define MAX_INT 0x7fffffff
#define MAX_LL 0x7fffffffffffffff
#define MAX(X,Y) ((X) > (Y) ? (X) : (Y))
#define MIN(X,Y) ((X) < (Y) ? (X) : (Y))

int main(){
//  fstream fin("C:\\Users\\Administrator\\Desktop\\in.txt",ios::in);

const char s[]="XHUSTACM";
LL l,r;
while(cin>>l>>r){
LL i=0,len=strlen(s);
LL j=(l-1)%len;
for(i=0; i<r-l+1; i++){
printf("%c",s[j]);
j=(j+1)%len;
}
printf("\n");
}

//  fin.close();
return 0;
}
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