#include <bits/stdc++.h>
using namespace std;
int a[4];
int main()
{
for (int i = 0; i < 4; i++)
scanf("%d", &a[i]);
sort(a, a + 4);
printf("%d %d %d", a[3] - a[0], a[3] - a[1], a[3] - a[2]);
} 如果只有两种数字并且差为偶数，除2
#include <bits/stdc++.h>
using namespace std;
map<int, int>mp;
int flag = -1;
int main()
{
int n, t;
scanf("%d", &n);
for (int i = 0; i < n; i++)
{
scanf("%d", &t);
mp[t]++;
}
vector<int>v;
for (auto it = mp.begin(); it != mp.end(); it++)
v.push_back(it->first);
if (v[0] == v[v.size() - 1])
{
printf("0");
return 0;
}
int f = 0, s = 0;
int cnt = mp.size();
for (int i = 1; i < v.size(); i++)
{
if (v[i - 1] != v[i])
{
f = v[i] - v[i - 1];
s = v[i];
break;
}
}
for (int i = 0; i < v.size(); i++)
{
if (v[i] != s)
{
if (v[i] < s && v[i] + f == s)
;
else if (v[i] > s && v[i] - f == s)
;
else
{
printf("-1");
return 0;
}
}
}
if (cnt == 2 && f % 2 == 0)
f /= 2;
printf("%d", f);
} 先把一个循环搞出来，然后算出剩下的，然后暴力模拟从周一到周日开始，跑循环
#include <bits/stdc++.h>
using namespace std;
int a[3], tt[3];
int eat[7] = { 0,1,2,0,2,1,0 };
int main()
{
scanf("%d%d%d", &a[0], &a[1], &a[2]);
int flag = 1e9;
flag = min(flag, a[0] / 3);
flag = min(flag, a[1] / 2);
flag = min(flag, a[2] / 2);
a[0] -= flag * 3;
a[1] -= flag * 2;
a[2] -= flag * 2;
int ans = flag * 7;
int temp = 0;
for (int j = 0; j < 7; j++)
{
tt[0] = a[0], tt[1] = a[1], tt[2] = a[2];
int t = 0;
for (int i = j; i < 49; i++)
{
i %= 7;
if (tt[eat[i]] > 0)
tt[eat[i]]--;
else
break;
t++;
}
temp = max(temp, t);
}
printf("%d", temp + ans);
} 如果蓄电池的电量满了，那么即使实在太阳下，我们也选择蓄电池
#include <bits/stdc++.h>
using namespace std;
int s[200005];
int main()
{
int n, b, a, maxn;
scanf("%d%d%d", &n, &b, &a);
maxn = a;
for (int i = 1; i <= n; i++)
scanf("%d", &s[i]);
for (int i = 1; i <= n; i++)
{
if (s[i] == 0)
{
if (a != 0)
a--;
else if (b != 0)
b--;
else
{
printf("%d", i - 1);
return 0;
}
}
else
{
if (a == maxn)
a--;
else if (b != 0)
{
b--;
if (a < maxn)
a++;
}
else if (a != 0)
a--;
else
{
printf("%d", i - 1);
return 0;
}
}
}
printf("%d", n);
} 删除的时候，将删除了的区间的左端点想上设为右端点，右端点向下设为左端点，然后，暴力他旁边的人的时候，跳着访问，249ms
#include <bits/stdc++.h>
using namespace std;
bool book[200005];
map<int, int>mp;
int ans[200005];
int toup[200005], todown[200005];
struct node
{
int num;
int index;
}que[200005];
auto cmp = [](node q, node w) {
return q.num < w.num;
};
priority_queue<node, vector<node> ,decltype(cmp)>q(cmp);
vector<int>v1, v2;
int main()
{
int n, m, cnt = 0;
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++)
{
toup[i] = i + 1;
todown[i] = i - 1;
scanf("%d", &que[i].num);
que[i].index = i;
q.push(que[i]);
mp[que[i].index] = que[i].num;
}
while (!q.empty())
{
qwe:;
if (q.empty())
break;
node temp = q.top();
q.pop();
if (book[temp.num] == true)
goto qwe;
book[mp[temp.index]] = true;
if (cnt % 2 == 0)
{
int up = temp.index;
int down = temp.index;
v1.push_back(temp.index);
for (int i = 1; i <= m; i++)
{
while (down >= 1 && book[mp[down]] == true)
down = todown[down];
if (down >= 1 && book[mp[down]] == false)
{
v1.push_back(down);
book[mp[down]] = true;
}
while (up <= n && book[mp[up]] == true)
up = toup[up];
if (up <= n && book[mp[up]] == false)
{
v1.push_back(up);
book[mp[up]] = true;
}
}
toup[down] = up;
todown[up] = down;
}
else
{
int up = temp.index;
int down = temp.index;
v2.push_back(temp.index);
for (int i = 1; i <= m; i++)
{
while (down >= 1 && book[mp[down]] == true)
down = todown[down];
if (down >= 1 && book[mp[down]] == false)
{
v2.push_back(down);
book[mp[down]] = true;
}
while (up <= n && book[mp[up]] == true)
up = toup[up];
if (up <= n && book[mp[up]] == false)
{
v2.push_back(up);
book[mp[up]] = true;
}
}
toup[down] = up;
todown[up] = down;
}
cnt++;
}
for (int i = 0; i < v1.size(); i++)
ans[v1[i]] = 1;
for (int i = 0; i < v2.size(); i++)
ans[v2[i]] = 2;
for (int i = 1; i <= n; i++)
printf("%d", ans[i]);
} 简单DP，先将价格数组排序，枚举到选了K个的情况，如果此时有选了K个的优惠，将选K个的优惠和最前的优惠取最大转移就可以了，赛时没时间想了。被E搞了好久。
#include <bits/stdc++.h>
#define ll long long
using namespace std;
map<int, int>mp;
int a[200005];
ll dp[2005], pre[200005];
int main()
{
int n, m, k;
scanf("%d%d%d", &n, &m, &k);
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]);
while (m--)
{
int a, b;
scanf("%d%d", &a, &b);
mp[a] = max(mp[a], b);
}
sort(a + 1, a + 1 + n);
for (int i = 1; i <= n; i++)
pre[i] = pre[i - 1] + a[i];
for (int i = 1; i <= k; i++)
{
dp[i] = dp[i - 1] + a[i];
for (auto it = mp.begin(); it != mp.end(); it++)
{
if (it->first > i)
break;
dp[i] = min(dp[i], dp[i - it->first] + pre[i] - pre[i - it->first + it->second]);
}
}
printf("%lld", dp[k]);
}
找有公因子的最小的两个数的lcm的最小值
1、首先枚举公因子 
2、枚举含有公因子的数 
3、复杂度
#include <bits/stdc++.h>
#define ll long long
using namespace std;
vector<int>v[10000005];
int s[1000005];
int main()
{
int n, l, r;
scanf("%d", &n);
for (int i = 1; i <= n; i++)
{
scanf("%d", &s[i]);
v[s[i]].push_back(i);
}
ll ans = 1e18;
for (int i = 1; i <= 100000000; i++)
{
int a = 0, b = 0;
for (int j = i; j <= 10000000; j += i)
{
for (int k = 0; k < v[j].size(); k++)
{
if (a == 0)
a = v[j][k];
else if (b == 0)
{
b = v[j][k];
break;
}
}
if (b != 0)
break;
}
if (b != 0)
{
if (ans > 1LL * s[a] * s[b] / i)
{
ans = 1LL * s[a] * s[b] / i;
l = a;
r = b;
}
}
}
if (l > r)
swap(l, r);
printf("%d %d", l, r);
}
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