HDU 5455(字符串)

最新推荐文章于 2019-07-04 10:28:42 发布

真的好难

最新推荐文章于 2019-07-04 10:28:42 发布

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分类专栏：水题字符串 acm

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本文探讨了一种名为FangFang的字符串序列，并提出了一系列基于该序列的算法挑战。任务要求通过最少数量的序列元素拼接给定的字符串。

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Fang Fang

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)

Total Submission(s): 849 Accepted Submission(s): 350

Problem Description

Fang Fang says she wants to be remembered.
I promise her. We define the sequence

F of strings.

F0 = ‘‘f",

F1 = ‘‘ff",

F2 = ‘‘cff",

Fn = Fn−1 + ‘‘f", for n > 2
Write down a serenade as a lowercase string

S in a circle, in a loop that never ends.
Spell the serenade using the minimum number of strings in

F, or nothing could be done but put her away in cold wilderness.

Input

An positive integer

T, indicating there are

T test cases.
Following are

T lines, each line contains an string

S as introduced above.
The total length of strings for all test cases would not be larger than

106.

Output

The output contains exactly

T lines.
For each test case, if one can not spell the serenade by using the strings in

F, output

−1. Otherwise, output the minimum number of strings in

F to split

Saccording to aforementioned rules. Repetitive strings should be counted repeatedly.

Sample Input

8
ffcfffcffcff
cffcfff
cffcff
cffcf
ffffcffcfff
cffcfffcffffcfffff
cff
cffc

Sample Output

Case #1: 3
Case #2: 2
Case #3: 2
Case #4: -1
Case #5: 2
Case #6: 4
Case #7: 1
Case #8: -1
Hint
Shift the string in the first test case, we will get the string "cffffcfffcff"
and it can be split into "cffff", "cfff" and "cff".

#include <iostream>
#include <stdio.h>
#include <string>
#include <string.h>
#include <cmath>
#include <algorithm>
#include <map>
#include <set>
#include <queue>
using namespace std;
const int inf=(1<<30);
char s[2000000];
int main()
{
    int t,cnt1=1;
    scanf("%d\n",&t);
    while(t--)
    {
        scanf("%s",s);
        int len=strlen(s);
        int t=len;
        int b=len;
        int cnt,mark;
        cnt=mark=0;
        for(int i=0;i<len;i++)
        {
            if(s[i]=='f')
                s[b++]='f';
            if(s[i]=='c')
                break;
        }
        s[b]='\0';
        int len1=strlen(s);
        printf("Case #%d: ",cnt1++);
        if(t*2==b)
        {
            if(len%2==0)
                printf("%d\n",len/2);
            else
                printf("%d\n",len/2+1);
            continue;
        }
        for(int i=0;i<len;i++)
        {
            if(s[i]=='c')
                cnt++;
            if(s[i]!='f'&&s[i]!='c')
                mark=1;
            if(s[i]=='c'&&s[i+1]!='f')
                mark=1;
            if(s[i]=='c'&&s[i+2]!='f')
                mark=1;
            if(s[len-1]=='c'&&s[0]!='f')
                mark=1;
            if(s[len-1]=='c'&&s[1]!='f')
                mark=1;
        }
        if(mark)
            printf("-1\n");
        else
            printf("%d\n",cnt);
    }
    return 0;
}