A

贪心,由于基地可以重复销毁，所以找出找出每架战机可轰炸基地的最大价值，排序之后贪心找一下即可

#include<iostream>
#include<algorithm>
using namespace std;
const int N = 1e6 + 15;

struct node{
    int d, v;
}no[N];
bool cmp(node a, node b){
    if(a.d == b.d)
        return a.v < b.v;
    else
        return a.d < b.d;
}
int a[N];
int main()
{
    int n, m; cin >> n >> m;
    for(int i = 0; i < n; i++)
        cin >> a[i];
    for(int i = 0; i < m; i++)
        cin >> no[i].d;
    for(int i = 0; i < m; i++)
        cin >> no[i].v;
    sort(a, a + n);
    sort(no, no + m, cmp);
    int ans = 0, p = 0;
    int maxd = 0;
    for(int i = 0; i < n; i++){
        while(p < m && a[i] > no[p].d){
            maxd = max(maxd, no[p].v);
            p++;
        }
        ans += maxd;
    }
    cout << ans << endl;
    return 0;
}

E

对比两个人答案的不同，如果不同答案大于k，则判定不同的k道题都是对的，num - k道是错的；如果不同答案小于k，则判定不同的都是对的，k - num道是错的

#include <iostream>
using namespace std;
const int N = 1005 + 15;
int a[N], b[N];
int main()
{
    int n, k;
    cin >> n >> k;
    for(int i = 0; i < n; i++)
        cin >> a[i];
    int num = 0;
    for(int i = 0; i < n; i++){
        int x; cin >> x;
        if(x != a[i])
            num++;
    }
    cout << n - (max(num, k) - min(num, k)) << endl;

    return 0;
}

G

只能出现前导0，所以有 n + 1种方案
举个例子即可n = 5
1 1 1 1 1
0 1 1 1 1
0 0 1 1 1
0 0 0 1 1
0 0 0 0 1
0 0 0 0 0

#include <iostream>
using namespace std;
int main()
{
    int n; cin >> n;
    cout << n + 1<< endl;
    return 0; 
}

H

两圆相交，注意：相切也算相交，同时注意double的精度问题

#include<iostream>
using namespace std;
#define ll long long
int main()
{
    int t; cin >> t;
    while(t--) {
        ll x, y, r;
        ll x1, y1, r1;
        cin >> x >> y >> r >> x1 >> y1 >> r1;
        ll dis = (y1 - y) * (y1 - y) + (x1 - x) * (x1 - x);

        ll rr = (r + r1) * (r + r1);
        ll rrr = ((max(r, r1) - min(r, r1))) * ((max(r, r1) - min(r, r1)));
        if (rrr <= dis && dis <= rr)
            cout << "YES" << endl;
        else
            cout << "NO"<< endl;
    }
    return 0;
}