A
#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
int main()
{
int n;
cin >> n;
string a;
cin >> a;
int one = ;
int zero = ;
for (int i = ; i < a.size(); i++)
{
if (a[i] == '')
{
one++;
}
else
{
zero++;
}
}
while (one >= )
{
one--;
}
for (int i = ; i <= one; i++)
{
cout << ;
}
for (int i = ; i <= zero; i++)
{
cout << ;
}
cout << endl;
}
B
模拟题 注意细节
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
int main()
{
ll n, m, k;
cin >> n >> m >> k;
ll ansx, ansy;
if (k <= n - )
{
cout << k + << " " << << endl;
return ;
}
if (k <= n + m - )
{
k -= n - ;
cout << n << " " << + k << endl;
return ;
}
k -= n + m - ;
//cout << "k " << k << endl;
ll duce = k / (m - );
//cout << "duce " << duce << endl;
int flag = ;
if (1LL * duce * (m - ) == k)
{
flag = ;
}
ll cha;
cha = k - 1LL * duce * (m - );
//cout << "cha " << cha << endl;
if (1LL * duce * (m - ) < k)
{
cha--;
duce++;
}
//cout << "cha " << cha << endl;
//cout << "duce " << duce << endl;
ansy = n - duce;
if ((duce + flag) % )
{
ansx = m - cha;
}
else
{
ansx = + cha;
}
cout << ansy << " " << ansx << endl;
}
C
先按L坐标从小到大排 L相等的时候R也从小到大排
从1到N遍历
假设存在一对i,j符合条件(i<j)且i与j之间的与i不满足与j也不满足 则[i].R>=[j].R 且 [k].R>=[i].R(i<k<j) 这样的话k与j也满足 假设不成立
所以我们知道只要比较相邻的两项就可以了
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
struct node
{
ll l, r;
ll index;
} num[];
bool operator <(node x, node y)
{
if (x.l == y.l)
{
return x.r < y.r;
}
return x.l < y.l;
}
int main()
{
int n;
cin >> n;
for (int i = ; i <= n; i++)
{
cin >> num[i].l >> num[i].r;
num[i].index = i;
}
sort(num + , num + + n);
for (int i = ; i <= n; i++)
{
if (num[i].l == num[i - ].l)
{
cout << num[i - ].index << " " << num[i].index << endl;
return ;
}
if (num[i].r <= num[i - ].r)
{
cout << num[i].index << " " << num[i - ].index << endl;
return ;
}
}
cout << - << " " << - << endl;
}
D
给N(N<=300)个数 Di (di<=1000) 要求你构建出一个无向图使得图满足有Dn+1个节点 这些节点的度数形成的集合为给你的数列
E
给你N,A,B N个生物 每个生物有HP和DMG两个属性 你可以进行A次操作每次使得一个生物的HP翻倍 B次操作每次使得一个生物的DMG=HP
问你最后这些生物最大的DMG和是多少
贪心 枚举情况思考很容易可以想到A次操作要全部放在一个生物上施展 接下来就是怎么用B和怎么算总DMG的问题了
计算CHA=HP-DMG B操作肯定是用在选择一个生物操作A后的前B大的非负CHA生物
有两种情况
1.当前选择操作A的生物是前B大的 2.当前选择操作A的不是前B大的
分类讨论就行了
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll sum = ;
ll ansernow;
ll anser = ;
ll add = ;
struct node
{
ll hp;
ll dmg;
ll cha;
} crea[];
bool operator <(node a, node b)
{
return a.cha > b.cha;
}
int main()
{
ll n, a, b;
cin >> n >> a >> b;
b = min(b, n);
for (int i = ; i <= n; i++)
{
cin >> crea[i].hp >> crea[i].dmg;
sum += crea[i].dmg;
crea[i].cha = crea[i].hp - crea[i].dmg;
}
sort(crea + , crea + + n);
if (b == )
{
cout << sum << endl;
return ;
}
ll bi = crea[b].cha;
for (int i = ; i <= b; i++)
{
//cout<<crea[i].cha<<" ";
if (crea[i].cha > )
{
add += crea[i].cha;
}
}
//cout<<endl;
//cout << "add:" << add << endl;
anser = add + sum;
if (a == )
{
cout << anser << endl;
return ;
}
ll chanow;
ll yuan;
ll cnt;
for(int i=;i<=b;i++)
{
ansernow=sum+add-max(0LL,crea[i].cha)+max(0LL,crea[i].hp*(1LL<<a)-crea[i].dmg);
anser=max(anser,ansernow);
}
for (int i = b+; i <= n; i++)
{
ansernow=sum+add-max(0LL,crea[b].cha)+max(0LL,crea[i].hp*(1LL<<a)-crea[i].dmg);
anser=max(anser,ansernow);
}
cout << anser << endl;
}