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  • AtCoder

    AtCoder - 3959
    https://vjudge.net/problem/1583855/origin
    求最长连续递增长度就行,答案是n-max(len)

    #include<iostream>
    #include<cstdio>
    #include<queue>
    #include<algorithm>
    #include<cmath>
    #include<ctime>
    #include<set>
    #include<map>
    #include<stack>
    #include<cstring>
    #define inf 2147483647
    #define ls rt<<1
    #define rs rt<<1|1
    #define lson ls,nl,mid,l,r
    #define rson rs,mid+1,nr,l,r
    #define N 1000010
    #define For(i,a,b) for(int i=a;i<=b;i++)
    #define p(a) putchar(a)
    #define g() getchar()
    
    using namespace std;
    int n;
    int a[N],b[N];
    int ans=1;
    void in(int &x){
        int y=1;
        char c=g();x=0;
        while(c<'0'||c>'9'){
            if(c=='-')y=-1;
            c=g();
        }
        while(c<='9'&&c>='0'){
            x=(x<<1)+(x<<3)+c-'0';c=g();
        }
        x*=y;
    }
    void o(int x){
        if(x<0){
            p('-');
            x=-x;
        }
        if(x>9)o(x/10);
        p(x%10+'0');
    }
    int main(){
        in(n);
        For(i,1,n)
            in(a[i]);
        b[a[1]]=1;
        For(i,2,n){
            b[a[i]]=b[a[i]-1]+1;
            ans=max(ans,b[a[i]]);
        }
        o(n-ans);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/war1111/p/11215285.html
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