NOIP 模拟 $92\; \rm 优美的旋律$

题解 \(by\;zj\varphi\)

先枚举起点，然后再枚举循环节长度，再暴力向后一节一节递推。

复杂度时合法的，外层有一个 \(n\)，内部复杂度为 \(n+\frac{n}{2}+\frac{n}{3}...+\frac{n}{n}\) 约为 \(n\ln n\)，总复杂度为 \(n^2\ln n\)。

Code

#include<bits/stdc++.h>
#define ri signed
#define pd(i) ++i
#define bq(i) --i
#define func(x) std::function<x>
namespace IO{
    char buf[1<<21],*p1=buf,*p2=buf;
    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++
    #define debug1(x) std::cerr << #x"=" << x << ' '
    #define debug2(x) std::cerr << #x"=" << x << std::endl
    #define Debug(x) assert(x)
    struct nanfeng_stream{
        template<typename T>inline nanfeng_stream &operator>>(T &x) {
            bool f=false;x=0;char ch=gc();
            while(!isdigit(ch)) f|=ch=='-',ch=gc();
            while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=gc();
            return x=f?-x:x,*this;
        }
    }cin;
}
using IO::cin;
namespace nanfeng{
    #define FI FILE *IN
    #define FO FILE *OUT
    template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
    template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
    using ull=unsigned long long;
    static const int N=3e3+7,P=13331;
    int mx[N],a,b,n;
    char s[N];
    ull hs[N],p[N],ans;
    inline int main() {
        FI=freopen("melody.in","r",stdin);
        FO=freopen("melody.out","w",stdout);
        scanf("%d%d%s",&a,&b,s+1);
        n=strlen(s+1);
        p[0]=1;
        for (ri i(1);i<=n;pd(i)) {
            hs[i]=hs[i-1]*P+(ull)(s[i]-'a'+1);
            p[i]=p[i-1]*P;
            mx[i]=1;
        }
        for (ri i(1);i<=n;pd(i)) {
            const int bs=n-i+1,len=bs>>1;
            for (ri j(1);j<=len;pd(j)) {
                if (mx[j]>=bs/j) continue;
                const ull on=p[j],jud=hs[i+j-1]-hs[i-1]*on;
                int nw=bs/j;
                for (ri k(2),l;(l=k*j)<=bs;pd(k))
                    if (hs[i+l-1]-hs[i+l-j-1]*on!=jud) {nw=k-1;break;}
                if (mx[j]<nw) ans=cmax(ans,1ull*a*j+1ull*b*nw),mx[j]=nw;
            }
        }
        printf("%llu\n",ans);
        return 0;
    }
}
int main() {return nanfeng::main();}