Gym

参考https://vjudge.net/solution/9596160和http://blog.csdn.net/prolightsfxjh/article/details/76099716

题意：给出字符串A和B，字符串A大小写不区分，字符串B由小写字母构成，构成AB的每种字母的个数相同，询问是否能把B切成3份，重构出字符串A，如果能则输出那三份子串，并按照A的顺序输出。

哈希+枚举

O(n^2)的枚举字符串B 的0~i-1 为第一段，i~j-1 为第二段，j~n-1为第三锻，

然后每次借用这三段的哈希值去和A串匹配，这三段有A(3,3) == 6种排列方式，所以枚举这6种排列，

之后借用它们的哈希值判断能否构造出字符串A。

复杂度 O(n^2)

可是我还是看不太懂这些地方的hash的用法。。。

#include<algorithm>#include<vector>#include<cstring>#include<string>#include<iomanip>#include<cstdio>#include<stack>#include<iostream>#include<map>#include<set>#include<queue>#include<cmath>#include<strstream>using namespace std;#define sf scanf#define pf printf#define mem(a,b) memset(a,b,sizeof(a));#define rep(i,a,b) for(int i=(a);i<=(b);++i)#define MP make_pair#define N 1000010#define M 200020#define ULL unsigned long long#define LL long long#define inf 0x3f3f3f3f//printf("\n%.2lf sec \n", (double)clock() / CLOCKS_PER_SEC);const int P=29;const int maxn=100000;ULL haha[N],TS[N];char aa[N],bb[N];ULL goal;int n;void print(char *s,int l,int r){    for(int i=l;i<=r;++i)pf("%c",s[i]);    puts("");}int main(){    TS[0]=1;    for(int i=1;i<=maxn;++i)TS[i]=TS[i-1]*P;    sf("%s",aa);    sf("%s",bb);    n=strlen(aa);    if(n<3){puts("NO");return 0;}    for(int i=0;i<n;++i){ goal=goal*P+aa[i]-'a'; }    for(int i=0;i<n;++i){haha[i+1]=haha[i]*P+bb[i]-'a'; }    for(int i=1;i<=n;++i){        for(int k=i+1;k<=n;++k){            ULL ha1=haha[i];            ULL ha2=haha[k]-haha[i]*TS[k-i];            ULL ha3=haha[n]-haha[k]*TS[n-k];            //123            if(ha1*TS[n-i]+ha2*TS[n-k]+ha3==goal){                puts("YES");                print(bb,0,i-1),print(bb,i,k-1),print(bb,k,n-1);                return 0;            }            //132            if(ha1*TS[n-i]+ha3*TS[k-i]+ha2==goal){                puts("YES");                print(bb,0,i-1);print(bb,k,n-1),print(bb,i,k-1);                return 0;            }            //213            if(ha2*TS[n-k+i]+ha1*TS[n-k]+ha3==goal){                puts("YES");                print(bb,i,k-1);print(bb,0,i-1);print(bb,k,n-1);                return 0;            }            //231            if(ha2*TS[n-k+i]+ha3*TS[i]+ha1==goal){                puts("YES");                print(bb,i,k-1),print(bb,k,n-1),print(bb,0,i-1);                return 0;            }            //321            //312        }    }    puts("NO");}