NOIP 模拟 $94\; \rm 叁仟柒佰万$

题解 \(by\;zj\varphi\)

性质：最终有贡献的方案的 \(mex\) 一定是全局 \(mex\)。

发现当右端点固定后，左端点 \(mex\) 值是单调的，所以双指针扫一下即可。

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,*p2;
    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++
    #define dg1(x) std::cerr << #x"=" << x << ' '
    #define dg2(x) std::cerr << #x"=" << x << std::endl
    #define Dg(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;}
    static const int N=3e5+7e4+7,MOD=1e9+7;
    int a[N*100],sum[N],dp[N*100],T,n;
    inline int main() {
        FI=freopen("clods.in","r",stdin);
        FO=freopen("clods.out","w",stdout);
        cin >> T;
        for (ri z(1);z<=T;pd(z)) {
            cin >> n;
            if (n==37000000) {
                int X,Y;
                cin >> X >> Y;
                const int x=X,y=Y;
                for (ri i(2);i<=n;pd(i)) a[i]=(1ll*a[i-1]*x+y+i)&262143; 
            } else for (ri i(1);i<=n;pd(i)) cin >> a[i];
            int amx=0,bg,pnt=0,l=1,r=1,su,mx=0;
            for (ri i(1);i<=n;pd(i)) {
                mx=cmax(mx,a[i]);
                ++sum[a[i]];
                while(sum[amx]) ++amx;
            }
            memset(sum,0,sizeof(int)*(mx+1));
            for (ri i(1);i<=n;pd(i)) {
                ++sum[a[i]];
                while(sum[pnt]) ++pnt;
                if (pnt==amx) {bg=i;break;}
            }
            su=dp[0]=dp[bg]=1;
            for (ri i(bg+1);i<=n;pd(i)) {
                ++sum[a[i]];
                while(r<i&&(sum[a[r]]>1||a[r]>amx)) {
                    if (r>=bg) su+=dp[r];
                    --sum[a[r++]];
                    if (su>=MOD) su-=MOD;
                }
                dp[i]=su;
            }
            printf("%d\n",dp[n]);
            if (z==T) break;
            memset(sum,0,sizeof(int)*(mx+1));
            memset(dp,0,sizeof(int)*(n+1));
        }
        return 0;
    }
}
int main() {return nanfeng::main();}