Codeforces Round #637 (Div. 2)

题目链接：http://codeforces.com/contest/1341

A

思路:判断n*(a-b)即最小的是否大于最大的(c+d),和n*(a+b)是否小于最小的(c-d)即可，其余的都是满足条件的

//-------------------------------------------------
//Created by HanJinyu
//Created Time :四  4/23 22:43:12 2020
//File Name :637A.cpp
//-------------------------------------------------

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
typedef double db;
typedef long long ll;
const int maxn = 200005;

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,a,b,c,d;
        scanf("%d%d%d%d%d",&n,&a,&b,&c,&d);
        int z=n*(a-b),y=n*(a+b);
        int zz=(c-d),yy=(c+d);
        if(z>yy||y<zz)
            printf("No\n");
        else
            printf("Yes\n");

    }
 
     return 0;
}

B

思路：当a[i]>a[i-1]&&a[i]>a[i+1]时，用数组标记一下，然后求出前缀和，记录当前共有几个峰，用b[i-1]-b[i-k+1]即能筛选出最大峰数的段，那么l就是i-k+1了。由于是被分成的段数，因此要峰数加1

//-------------------------------------------------
//Created by HanJinyu
//Created Time :四  4/23 22:43:12 2020
//File Name :637Bcpp
//-------------------------------------------------

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
typedef double db;
typedef long long ll;
const int maxn = 200005;
int main()
{
    freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int k,n;
        scanf("%d%d",&n,&k);
        ll a[maxn],b[maxn],c[maxn];
        memset(b,0,sizeof(b));
        memset(c,0,sizeof(c));
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&a[i]);
        }
       // b[1]=0;c[1]=0;
        for(int i=2;i<=n-1;i++)
        {
            if(a[i]>a[i-1]&&a[i]>a[i+1])
                b[i]=1;
            else
                b[i]=0;
        }
        for(int i=1;i<=n;i++)
                b[i]+=b[i-1];
        int res=0,l=1;
        for(int i=k;i<=n;i++)
        {
            if(b[i-1]-b[i-k+1]>res)
            {
                res=b[i-1]-b[i-k+1];
                l=i-k+1;
            }
        }
        printf("%d %d\n",res+1,l);
    }
 
     return 0;
}

C

思路：额，据说看样例就能发现规律：只要是Yes的数组里，左边那个数比右边那个数小的差值都是1，是No的存在小的和相邻右边的差值不为1.。。。

//-------------------------------------------------
//Created by HanJinyu
//Created Time :二  4/28 22:47:43 2020
//File Name :637C.cpp
//-------------------------------------------------

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
typedef double db;
typedef long long ll;
const int maxn = 200005;

int main()
{
    // freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int T;
    scanf("%d",&T);
    while(T--){

        int n;
        scanf("%d",&n);
        int a[maxn];
        for(int i=0;i<n;i++)
        {

            scanf("%d",&a[i]);
        }
        bool flag=false;
        for(int i=0;i<n-1;i++)
        {

            if(a[i]<a[i+1])
            {

                if(a[i]!=a[i+1]-1)
                    {

                        flag=true;
                        break;
                    }
            }
        }
        if(!flag||n==1)
            printf("Yes\n");
        else
            printf("No\n");
    }
 
     return 0;
}