luogu 1360 阵容均衡(前缀和+差分+hash)

要求一段最大的区间里每个能力的增长值是一样的。

我们首先求一遍前缀和，发现，如果区间内[l,r]每个能力的增长值是一样的话，那么前缀和[r]和[l-1]的差分也应该是一样的。

那么我们把前缀和的差分hash一下，排个序，求出hash值相同的且跨越最远的两个端点，就是答案了。

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-
# define MOD
# define INF
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<,l,mid
# define rch p<<|,mid+,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
void Out(int a) {
    if(a<) {putchar('-'); a=-a;}
    if(a>=) Out(a/);
    putchar(a%+'');
}
const int N=;
//Code begin...

int a[N], wei[N][];
struct Node{int hash, id;}node[N];

bool comp(Node a, Node b){
    if (a.hash!=b.hash) return a.hash<b.hash;
    return a.id<b.id;
}
int main ()
{
    int n, m, x;
    scanf("%d%d",&n,&m);
    FOR(i,,n) {
        scanf("%d",&x);
        int num=;
        while (x) wei[i][++num]=x%, x/=;
        FOR(j,,m) wei[i][j]+=wei[i-][j];
    }
    FOR(i,,n) {
        int val=;
        FOR(j,,m) val=val*+wei[i][j]-wei[i][j-];
        node[i].hash=val; node[i].id=i;
    }
    sort(node,node+n+,comp);
    int ans=, l, r;
    FOR(i,,n) {
        if (i==||node[i].hash!=node[i-].hash) l=node[i].id;
        if (i==n||node[i].hash!=node[i+].hash) r=node[i].id, ans=max(ans,r-l);
    }
    printf("%d\n",ans);
    return ;
}