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-7
# define MOD 1024523
# define INF 1000000000
# 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<<1,l,mid
# define rch p<<1|1,mid+1,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=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=100005;
//Code begin...

int a[N], wei[N][35];
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,1,n) {
        scanf("%d",&x);
        int num=0;
        while (x) wei[i][++num]=x%2, x/=2;
        FOR(j,1,m) wei[i][j]+=wei[i-1][j];
    }
    FOR(i,0,n) {
        int val=0;
        FOR(j,2,m) val=val*3+wei[i][j]-wei[i][j-1];
        node[i].hash=val; node[i].id=i;
    }
    sort(node,node+n+1,comp);
    int ans=0, l, r;
    FOR(i,0,n) {
        if (i==0||node[i].hash!=node[i-1].hash) l=node[i].id;
        if (i==n||node[i].hash!=node[i+1].hash) r=node[i].id, ans=max(ans,r-l);
    }
    printf("%d
",ans);
    return 0;
}