• 【Codeforces Round #424 (Div. 2) A】Unimodal Array


    Link:http://codeforces.com/contest/831/problem/A

    Description

    让你判断一个数列是不是这样一个数列:
    一开始是严格上升
    然后开始全都是一个数字
    然后又开始严格下降

    Solution

    枚举中间全部相同的是哪个数字;
    假设是i..j这一段
    则看看1..i是不是严格上升,以及j+1..n是不是严格下降;
    如果所有的这样的i..j都不满足则无解;
    找到一组则有解;

    NumberOf WA

    0

    Reviw

    找到最特殊的地方;
    抓住问题的重点!

    Code

    #include <bits/stdc++.h>
    using namespace std;
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    #define LL long long
    #define rep1(i,a,b) for (int i = a;i <= b;i++)
    #define rep2(i,a,b) for (int i = a;i >= b;i--)
    #define mp make_pair
    #define pb push_back
    #define fi first
    #define se second
    #define ms(x,y) memset(x,y,sizeof x)
    #define Open() freopen("F:\rush.txt","r",stdin)
    #define Close() ios::sync_with_stdio(0)
    
    typedef pair<int,int> pii;
    typedef pair<LL,LL> pll;
    typedef set<int> myset;
    
    const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
    const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
    const double pi = acos(-1.0);
    const int N = 2e3;
    
    struct abc{
        int x,id;
    };
    
    int k,n,a[N+100],b[N+100],pre[N+100],ans,tot;
    map <int,int> dic;
    abc c[N*N+10];
    
    bool cmp(abc a,abc b){
        return a.x < b.x;
    }
    
    int main(){
        //Open();
        //Close();
        scanf("%d%d",&k,&n);
        rep1(i,1,k)
            scanf("%d",&a[i]);
        pre[0] = 0;
        rep1(i,1,k)
            pre[i] = pre[i-1] + a[i];
        rep1(i,1,n)
            scanf("%d",&b[i]);
        rep1(i,1,k){
            rep1(j,1,n){
                int x = b[j]-pre[i];
                tot++;
                c[tot].x = x,c[tot].id = j;
            }
        }
        sort(c+1,c+1+tot,cmp);
        rep1(i,1,tot){
            int num = n;
            int j = i;
            while (j+1<=tot && c[j+1].x==c[i].x) j++;
            rep1(k,i,j){
                if (dic[c[k].id]!=i){
                    num--;
                    dic[c[k].id] = i;
                }
            }
            if (num==0) ans++;
            i = j;
        }
        printf("%d
    ",ans);
        return 0;
    }
    
    /*
        写完之后,明确每一步的作用
    */
    
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  • 原文地址:https://www.cnblogs.com/AWCXV/p/7626195.html
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