[BalticOI2014]Friends/[BZOJ4287]新三个和尚

[BalticOI2014]Friends/[BZOJ4287]新三个和尚

题目大意：

一个字符串\(A\)，将\(A\)复制一遍并在任意位置插入一个新字符得到\(B\)。给出\(B(|B|\le2\times10^6,|\Sigma|\le26)\)，求\(A\)是否存在、是否唯一。若唯一，则求出\(A\)。

思路：

字符串哈希。

源代码：

#include<cstdio>
#include<cctype>
inline int getint() {
    register char ch;
    while(!isdigit(ch=getchar()));
    register int x=ch^'0';
    while(isdigit(ch=getchar())) x=(((x<<2)+x)<<1)+(ch^'0');
    return x;
}
inline char getalpha() {
    register char ch;
    while(!isalpha(ch=getchar()));
    return ch;
}
typedef unsigned long long uint64;
const int N=2e6+1;
const uint64 base=31;
int s[N];
uint64 pwr[N],hash[N];
inline uint64 calc(const int &i,const int &j) {
    return hash[j]-hash[i-1]*pwr[j-i+1];
}
int main() {
    const int n=getint();
    if(n%2==0) {
        puts("NOT POSSIBLE");
        return 0;
    }
    for(register int i=pwr[0]=1;i<=n;i++) {
        s[i]=getalpha();
        pwr[i]=pwr[i-1]*base;
        hash[i]=hash[i-1]*base+s[i]-'A'+1;
    }
    uint64 last=0;
    int pos;
    for(register int i=1;i<=n/2;i++) {
        if(calc(1,i-1)*pwr[n/2-i+1]+calc(i+1,n/2+1)==calc(n/2+2,n)) {
            if(last!=0&&calc(n/2+2,n)!=last) {
                puts("NOT UNIQUE");
                return 0;
            }
            last=calc(n/2+2,n);
            pos=i;
        }
    }
    if(calc(1,n/2)==calc(n/2+2,n)) {
        if(last!=0&&calc(1,n/2)!=last) {
            puts("NOT UNIQUE");
            return 0;
        }
        last=calc(1,n/2);
        pos=n/2+1;
    }
    for(register int i=n/2+2;i<=n;i++) {
        if(calc(1,n/2)==calc(n/2+1,i-1)*pwr[n-i]+calc(i+1,n)) {
            if(last!=0&&calc(1,n/2)!=last) {
                puts("NOT UNIQUE");
                return 0;
            }
            last=calc(1,n/2);
            pos=i;
        }
    }
    if(!last) {
        puts("NOT POSSIBLE");
        return 0;
    }
    if(pos<=n/2) {
        for(register int i=n/2+2;i<=n;i++) putchar(s[i]);
    } else {
        for(register int i=1;i<=n/2;i++) putchar(s[i]);
    }
    putchar('\n');
    return 0;
}