2022年3月20日22:13:05

4376. 数圈圈

1.【题目描述】

输入样例1：
11
输出样例1：
2

输入样例2：
14
输出样例2：
0

2.【题解】

进制转换+字符串遍历

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
const int inf=0x3f3f3f3f;
const int maxn=1e6+5;
string func(int n){
   
	string s;
	int t = n / 16;
	int l = n % 16;
	if (t > 0)
		s += func(t);
	if (l < 10)
		s += (l + '0');
	else
		s += ('A' + l - 10);
	return s;
}
void solve(){
   
    int n;
    cin >> n;
    string ss = func(n);
    int sum = 0;
    for (int i = 0; i < ss.size();i++){
   
        if(ss[i]=='0'||ss[i]=='4'||ss[i]=='6'||ss[i]=='9'||ss[i]=='A'||ss[i]=='D'){
   
            sum += 1;
        }else if(ss[i]=='8'||ss[i]=='B'){
   
            sum += 2;
        }
    }
    cout << sum;
}

int main(){
   
    solve();
    return 0;
}

4377. 农田灌溉

1.【题目描述】

输入样例：
3
5 1
3
3 3
1 2 3
4 1
1
输出样例：
3
1
4

2.【题解】

暴力按题意模拟，找到最大的距离
ans=max(两喷嘴之间的最大值，靠边喷嘴距边的最大值）

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
const int inf=0x3f3f3f3f;
const int maxn=1e6+5;
void solve(){
   
    int n, k;
    cin >> n >> k;
    int sa[205] = {
   0};
    int llen,rlen;
    for (int i = 0; i < k;i++){
   
        int x;
        cin >> x;
        if(i==0)llen=x;
        if(i==k-1)rlen=n-x+1;
        sa[x] = 1;
    }
    sa[0] = 1;
    sa[n+1] = 1;
    int maxx = -1;
    int l, r;
    for (int i = 0; i <= n+1;i++){
   
        if(sa[i]){
   
            int cnt = 0;
            for (int j = i + 1; j <= n+1;j++){
   
                if(sa[j]){
   
                    cnt = j - i;
                    if(maxx<cnt){
   
                        l = i;
                        r = j;
                        maxx = cnt;
                    }    
                    break;
                }
            }
        }
    }
    //cout << l << " " << r << " " << maxx<<endl;
    int m=max(llen,rlen);//靠边喷嘴距边的最大值
    if(ceil(maxx/2+0.5)>m){
   
        m=ceil(maxx/2+0.5);//两喷嘴之间的最大值
    }
    if(n==k){
   
        cout << 1 << endl;
    }else if(l==1&&r==n){
   
        cout << m << endl;
    }else if (r==n+1||l==0){
   
        cout << maxx << endl;
    }else{
   
        cout << m << endl;
    }
}

int main(){
   
    ios::sync_with_stdio(0);
    int t;
    cin>>t;
    while(t--)
    solve();
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
const int inf=0x3f3f3f3f;
const int maxn=1e6+5;
void solve(){
   
    int n, k;
    cin >> n >> k;
    q[0] = 1;
    for (int i = 1; i < k + 1;i++)
        cin >> q[i];
    int res = q[1] - q[0] + 1;
    for (int i = 2; i < k + 1;i++)
        res = max(res, (q[i] - q[i - 1]) / 2 + 1);
    res = max(res, n - q[k] + 1);
    cout << res << endl;
}

int main(){
   
    ios::sync_with_stdio(0);
    int t;
    cin>>t;
    while(t--)
    solve();
    return 0;
}

4378. 选取数对

1.【题目描述】

输入样例1：
5 2 1
1 2 3 4 5
输出样例1：
9
输入样例2：
7 1 3
2 10 7 18 5 33 0
输出样例2：
61

2.【题解】

前缀和+01背包

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
const int inf=0x3f3f3f3f;
const int maxn=5020;
ll a[maxn];
ll sum[maxn], f[maxn][maxn];
int main(){
   
    int n, m, k;
    cin >> n >> m >> k;
    for (int i = 1; i <= n;i++) cin >> a[i];
    for (int i = 1; i <= m;i++) sum[1] += a[i];
    for (int i = 2; i + m - 1 <= n;i++)
        sum[i] = sum[i-1] - a[i-1] + a[i+m-1];
    for (int i = 1; i + m - 1 <= n;i++)
        for (int j = 0; j <= k;j++){
   
            f[i][j] = f[i-1][j];
            if (i >= m && j >= 1){
   
                f[i][j] = max(f[i][j], f[i-m][j-1] + sum[i]);
            }else if(j>=1){
   
                f[i][j] = max(f[i][j], sum[i]);
            }
        }
    cout << f[n-m+1][k] << endl;
}