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HDU 1358 Period

編輯：關於C++

Period

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3749 Accepted Submission(s): 1841

Problem Description For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as A^K , that is A concatenated K times, for some string A. Of course, we also want to know the period K.

Input The input file consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S. The second line contains the string S. The input file ends with a line, having the number zero on it.

Output For each test case, output “Test case #” and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.

Sample Input

3
aaa
12
aabaabaabaab
0

Sample Output

Test case #1
2 2
3 3

Test case #2
2 2
6 2
9 3
12 4

呃。。題意大意就是，從第二個位置開始，如果前面有相同的循環串，則輸出位置此時的位置下表，並且輸出循環串的個數。思路：跑個KMP，（只有某個位置的移動距離與當前位置的相差的值）為當前位置的1/m之一時，並且next數組裡面的值不為0時，才有循環串出現。（窩時這樣理解的，不知道對不對）。。上代碼。。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include
#include 
using namespace std;
#define mod 1000000007
#define inf 0x7f7f7f7f
int Next[1000005];
char s[1000005]; //看好題目，，數組開100W，我開了10WRE了好多次。。
int l;
void getNext()
{
    int i,j;
    i=0;
    j=-1;
    Next[i]=j;
    while(i>n,n)
    {
        scanf("%s",s);
        l=strlen(s);
        memset(Next,0,sizeof(Next));
        getNext();
        printf("Test case #%d\n",++Case);
        int k;
        // for(k=0;k


所以，求出