'只要坚持就会成功--2022.1.19'

题目链接: 懒惰的牛

题目代码:

#include<iostream>
#include<cstdio>

using namespace std;

const int N=3e6+10;
typedef long long ll;
int n,k;
ll a[N];

int main()
{
    cin>>n>>k;
    int r=0;
    for(int i=1;i<=n;i++){
        int  g,x;
        scanf("%d%d",&g,&x);
        r=max(r,x);
        a[max(0,x-k)]+=g,a[x+k+1]-=g;//差分 + 因为x大于等于0，所以可以用数组存;
    }                               // 另外如果草地范围大于草地到0的距离的话，左端可以直接从0开始加;
    ll res=0;
    for(int i=1;i<=r;i++){
        a[i]+=a[i-1];
        res=max(res,a[i]);
    }
    cout<<res<<endl;
    return 0;
}

题目链接: 金发姑娘和 N 头牛

题目代码: 

#include <iostream>
#include <cstring>
#include <algorithm>
#include <map>

using namespace std;

const int INF = 2e9;

int n, x, y, z;

int main()
{
    map<int, int> b;

    scanf("%d%d%d%d", &n, &x, &y, &z);
    for (int i = 0; i < n; i ++ )
    {
        int l, r;
        scanf("%d%d", &l, &r);
        b[-INF] += x;
        b[l] += y - x;
        b[r + 1] += z - y;
        b[INF] -= z;
    }

    int res = 0, sum = 0;
    for (auto& [k, v]: b)
    {
        sum += v;
        res = max(res, sum);
    }

    printf("%d\n", res);
    return 0;
}

题目链接: 贝茜放慢脚步

题目代码: 

#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
//归并排序;
using namespace std;

int n;
vector<int> a, b;

int main()
{
    scanf("%d", &n);
    
    char str[2];
    int x;
    
    while(n --)
    {
        scanf("%s%d", str, &x);
        if(*str == 'T') a.push_back(x);
        else b.push_back(x);
    }
    
    b.push_back(1000);
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());
    
    double v = 1, t = 0, s = 0;
    int i = 0, j = 0;
    while(i < a.size() || j < b.size())
    {
        if(j == b.size() || i < a.size() && a[i] - t < (b[j] - s) * v)
        {
            s += (a[i] - t) / v;
            t = a[i];
            v ++;
            i ++;
        }
        
        else 
        {
            t += (b[j] - s) * v;
            s = b[j];
            v ++;
            j ++;
        }
    }
    
    printf("%.0lf\n", t);
    return 0;
}