hdu1711 hdu1686 hdu3336 hdu1358

hdu 1711

kmp算法模板题，没明白问什么这里下标一定要是1开始，0开始就错了，或许是我还有地方没想到只是过了样例而已，改为1过了

代码：

#include <iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<cmath>#include<map>#include<algorithm>using namespace std;const int LEN1=1000105;const int LEN2=10105;int s[LEN1], t[LEN2], next[LEN2]; void get_next(const int&m){    int i=1, j=0; next[1]=0;    while( i<=m )    {        if( 0==j || t[i]==t[j] )        {            ++i; ++j; next[i]=j;        }        else j=next[j];    }}int func(const int&n, const int&m){    int i=1, j=1;    while( i<=n && j<=m )    {        if( 0==j || s[i]==t[j] )        {            ++i; ++j;        }        else j=next[j];    }    if( j>m ) return i-m;    return -1;}int main(int argc, char *argv[]){    int T, n, m;    cin>>T;    while( T-- )    {        scanf("%d%d", &n, &m);        for(int i=1; i<=n; i++) scanf("%d", s+i);        for(int i=1; i<=m; i++) scanf("%d", t+i);        get_next(m);        printf("%d\n", func(n, m));    }    return 0;}

hdu 1686

和hdu 1711差不多，不过这里是求s串里最大有几个t串

代码：

#include <iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<cmath>#include<map>#include<algorithm>using namespace std;const int LEN1=1000105;const int LEN2=10105;char s[LEN1], t[LEN2];int next[LEN2]; void get_next(){int i=0, j=-1, m=strlen(t); next[0]=-1;while( i<m ){if( -1==j || t[i]==t[j] ){++i; ++j; next[i]=j;}else j=next[j];}}int kmp(){int i=0, j=0, n=strlen(s), m=strlen(t);int times=0;while( i<n ){if( -1==j || s[i]==t[j] ){++i; ++j;}else j=next[j];/*这里不是j=0;因为i到这里 s串的 i-next[j]+1 ~~~ i-1会和 t串的 0 ~~~ next[j]-1相同但不一定s[i]和t[0]相同，所以如果 j=0会减少最大子串的数目； */ if( j>=m ) times++, j=next[j];}return times;}int main(int argc, char *argv[]){int T;cin>>T;while( T-- ){scanf("%s", t);scanf("%s", s);get_next();printf("%d\n", kmp());}return 0;}

hdu1358:

思路：因为此题是一个子字符串循环得到的母串，所以没有特判；直接就是算后面的是前面的小字符串的多少倍，利用next数组，因为

next数组存放的是与前缀相同的最长的序列下标到了哪里；所以  i - next[i]  表示next[i]+1 ~ i 都是1 ~ next[i] 的字符串复制得到；所以做就行了

代码：

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<cmath>#include<map>#include<algorithm>using namespace std;const int LEN=1000005;const int INF=0x3f3f3f3f;const int MOD=10007;int next[LEN];  char s[LEN], t[LEN];int cns[LEN];int main(int argc, char *argv[]){int i, j, lt, cas=0;while( scanf("%d", <)!=EOF && lt ){getchar();scanf("%s", t+1);i=1; j=0; next[1]=0;while( i<=lt ){if( 0==j || t[i]==t[j] ){++i; ++j;next[i]=j;}else j=next[j];}for(i=1; i<=lt; i++) cout<<next[i]<<" "; cout<<endl;printf("Test case #%d\n", ++cas);for(i=3, lt++; i<=lt; i++){j=(i-1)/(i-next[i]);if( j>1 && (i-1)%(i-next[i])==0 )printf("%d %d\n", i-1, j);}printf("\n");}return 0;}

hdu3336

思路：

代码：

#include <iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<cmath>#include<map>#include<algorithm>using namespace std;const int LEN = 1000005;const int MOD=10007;char t[LEN];int next[LEN];int main(int argc, char *argv[]){    int n, i, j, T;    cin>>T;    while( T-- )    {        scanf("%d%s", &n, t);        i=0; j=-1; next[0]=-1;        while( i<n )        {            if( -1==j || t[i]==t[j] ) next[++i]=++j;            else j=next[j];        }        //for(i=0; i<n; i++) cout<<next[i]<<" ";        //cout<<endl;        int sum=0, pos;        for(i=1; i<=n; i++)        {            pos=i;            while( pos )            {                sum=(sum+1)%MOD;                pos=next[pos];            }        }        printf("%d\n", sum);    }    return 0;}