1.股票买卖 II
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
const int N =100010;
int a[N];
int n;
int main()
{
cin >> n;
for(int i = 1; i <= n; i++) cin >> a[i];
int res = 0;
for(int i = 1; i < n; i++)
{
int t = a[i+1] - a[i];//只要后面的比前面的大,就进行交易
if(t > 0) res += t;
}
cout << res <<endl;
return 0;
}
2. 糖果传递
ACWing122. 糖果传递
输入样例:
4
1
2
5
4
输出样例:
4
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
const int N = 1000010;
int n;
int a[N];
long long c[N];
int main()
{
cin >>n;
for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
long long sum = 0;
for(int i = 1; i <= n; i++) sum += a[i];
long long avg = sum/n;
for(int i = n; i > 1; i--)
{
c[i] = c[i+1]+avg - a[i];
}
c[1] = 0;
sort(c+1,c+n+1);
long long res = 0;
for(int i = 1; i <= n; i++) res += abs(c[i]-c[(i+1)/2]);
cout << res << endl;
return 0;
}
3. 雷达设备
ACWing112. 雷达设备
输入样例:
3 2
1 2
-3 1
2 1
输出样例:
2
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
const int N = 1010;
int n, d;
struct Segment//存线段
{
double l,r;//左右端点
bool operator<(const Segment& t)const
{
return r < t.r;
}
}seg[N];
int main()
{
cin >> n >> d;
bool fail = false;
for(int i = 0; i <n; i++)
{
int x,y;
cin >> x >> y;
if(y > d) fail = true;//无解
else
{
double len = sqrt(d*d - y*y);
seg[i] = {
x-len, x + len};
}
}
if(fail) puts("-1");//如果已经失败
else
{
sort(seg,seg+n);
int cnt = 0;
double last = -1e20;//最开始没有坐标,默认负无穷
for(int i = 0; i< n; i++)//从前往后依次枚举每个线段
if(last < seg[i].l)//如果上一个坐标的右边界小于当前坐标左边界
{
cnt++;//雷达数增加
last = seg[i].r;//更新last
}
cout << cnt << endl;
}
return 0;
}
4.付账问题
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
const int N = 500010;
int a[N];
double b[N];
int n;
double s;
int main()
{
cin >>n >> s;
for(int i= 0; i < n; i++) scanf("%d", &a[i]);
sort(a,a+n);
double res = 0,avg = s/n;
for(int i = 0; i <n; i++)
{
double cur = s/(n-i);
if(a[i] < cur) cur = a[i];//如果a[i]<平均值, 只能出a[i] ,否则直接出平均值
res += (cur - avg)*(cur - avg);
s -= cur;
}
printf("%.4lf\n",sqrt(res/n));
return 0;
}
5.乘积最大
Acwing1239. 乘积最大
输入样例1:
5 3
-100000
-10000
2
100000
10000
输出样例1:
999100009
输入样例2:
5 3
-100000
-100000
-2
-100000
-100000
输出样例2:
-999999829
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
const int N = 100010,mod = 1000000009;
int a[N];
int n, k;
int main()
{
cin >>n >>k;
int res = 1;
for(int i = 0; i <n; i++)
{
scanf("%d", &a[i]);
}
sort(a,a+n);
//如果k %2 == 0 答案必然是在左右两边各选一堆(绝对值前k大的数)
//如果k%2 != 0 把最大的数选掉,就变成了偶数的情况,k变为偶数,从左右两边往中间取即可
int l = 0, r = n-1;
int sign = 1;//符号
if(k%2)
{
res = a[r--];
k--;
if(res < 0) sign = -1;//如果结果为负
}
while(k)
{
LL x = (LL)a[l]*a[l+1],y = (LL)a[r-1]*a[r];
if(x *sign >y*sign)
{
res = x%mod * res%mod;
l += 2;
}
else
{
res = y%mod*res%mod;
r -= 2;
}
k -= 2;
}
cout << res << endl;
return 0;
}
6.后缀表达式
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 200010;
typedef long long LL;
int n,m,a[N];
int main()
{
scanf("%d%d", &n, &m);
int k = n+m+1;
for(int i = 0; i < k; i++) scanf("%d", &a[i]);
LL res = 0;
if(!m)//如果没有负号,求和
{
for(int i = 0; i < k; i++) res += a[i];
}
else//有负号,1~m,至少可选1个负号,加一个数减去就可以将负号改为正号,例如:b-(a0-a1-a2-a3....)
{
sort(a,a+k);
res = a[k-1]-a[0];//至少一个减号,最大值,减最小值
for(int i = 1; i <k-1; i++) res += abs(a[i]);
}
printf("%lld", res);
return 0;
}
7. 灵能传输
AcWing1248. 灵能传输
总体思路
该题实际上要求通过何种灵能传输可以使得该序列的最大值最小
而由前缀和可知 一个有序的前缀和序列 其max(s[i]-s[i-1])的最大值可以达到最小
参考这篇问章Ni
2.通过上述分析可以明确想要求得本题的最优解应使得所求序列尽量保持单调
通过画图可知一个有两个拐点的曲线重叠部分最小时 单调部分最多
而一个曲线符合下列情况时符合要求:
- 左端点小于右端点 即要求s[0]<s[n]
//所以在记录s0和sn的值的时候 需要进行一次特判
if(s0>sn)
swap(s0,sn);
- 2.极小值在极大值左边
该点要求我们在后续选点的时
应s[0]向左取 s[n]向右取 因为只有这样才能取得两边的极值
int l = 0, r = n;
for (int i = s0; i >= 0; i -= 2)
{
a[l ++ ] = s[i];
st[i] = true;
}
for (int i = sn; i <= n; i += 2)
{
a[r -- ] = s[i];
st[i] = true;
}
for (int i = 0; i <= n; i ++ )
if (!st[i])
a[l ++ ] = s[i];
- 这样以后就可以保证序列为f为重叠部分最小的前缀和序列
LL res = 0;
for (int i = 1; i <= n; i ++ ) res = max(res, abs(a[i] - a[i - 1]));
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 300010;
int n;
LL a[N], s[N];
bool st[N];
int main()
{
int T;
scanf("%d", &T);
while (T -- )
{
scanf("%d", &n);
s[0] = 0;
for (int i = 1; i <= n; i ++ )
{
scanf("%lld", &a[i]);
s[i] = s[i - 1] + a[i];
}
LL s0 = s[0], sn = s[n];
if (s0 > sn) swap(s0, sn);
sort(s, s + n + 1);
for (int i = 0; i <= n; i ++ )
if (s[i] == s0)
{
s0 = i;
break;
}
for (int i = n; i >= 0; i -- )
if (s[i] == sn)
{
sn = i;
break;
}
memset(st, 0, sizeof st);
int l = 0, r = n;
for (int i = s0; i >= 0; i -= 2)
{
a[l ++ ] = s[i];
st[i] = true;
}
for (int i = sn; i <= n; i += 2)
{
a[r -- ] = s[i];
st[i] = true;
}
for (int i = 0; i <= n; i ++ )
if (!st[i])
a[l ++ ] = s[i];
LL res = 0;
for (int i = 1; i <= n; i ++ ) res = max(res, abs(a[i] - a[i - 1]));
printf("%lld\n", res);
}
return 0;
}