菜鸡在这里做做笔记,各位进来的大佬我先 orz 了
B
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 2e5 + 10;
const int inf = 0x3f3f3f3f;
int n;
int main()
{
long long l, r, a;
cin >> n;
while (n--)
{
cin >> l >> r >> a;
if (a == 1)
{
cout << r << endl;
continue;
}
ll p = (r / a) * a - 1LL;//p为mod a后等于a-1的数
if (p < l)p = l;
ll ans = max(p / a + p % a, r / a + r % a);
cout << ans << endl;
}
return 0;
}
p = (r / a) * a - 1LL的原理
![](https://img-blog.csdnimg.cn/25b1e8f4dcad4ce687d26664e9ed4d1a.png?x-oss-process=image/watermark,type_d3F5LXplbmhlaQ,shadow_50,text_Q1NETiBAWENQUA==,size_20,color_FFFFFF,t_70,g_se,x_16)
C
本题为贪心算法,权重最高的前2n个节点权重累加,建立嵌套段系统时只需讲第i个位置和n+1-i位置连接起来即可,输出时使用该输出方法使得节点下标小的节点在前。
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 2e5 + 10;
const int inf = 0x3f3f3f3f;
struct node
{
ll loc, w;
ll idx;
}a[N],b[N];
bool cmp1(node x1, node x2)
{
return x1.w < x2.w;
}
bool cmp2(node x1, node x2)
{
return x1.loc < x2.loc;
}
void solve()
{
ll n, m, ans = 0;
cin >> n >> m;
for (ll i = 1; i <= m; i++)
{
cin >> a[i].loc >> a[i].w;
a[i].idx = i;
}
sort(a + 1, a + 1 + m, cmp1);
ll r = n * 2;
for (ll i = 1; i <= r; i++)
{
ans += a[i].w;
b[i] = a[i];
}
sort(b + 1, b + 1 + r, cmp2);
cout << ans << endl;
for (int i = 1; i <=n; i++)
cout << min(b[i].idx, b[r - i + 1].idx) << " " << max(b[i].idx, b[r + 1 - i].idx) << endl;
cout << endl;
}
int main()
{
int _;
cin >> _;
while (_--)
{
solve();
}
return 0;
}