# 9.11考试总结
细胞分裂
数学题目，因式分解后直接判断输入数据是否含有m1中分解出来的数，然后储存需要时间最大值中的最小值
#include<bits/stdc++.h>
#define open(s) freopen(s".in", "r", stdin);// freopen(s".out", "w", stdout);
#define IL inline
#define ull unsigned long long
#define ll long long
using namespace std;
int n, m1, m2;
int a;
int maxb, lenb;
int b[30010];
struct date
{
int su, add;
}d[30010];
int lend;
IL int read();
IL void cut(int);
IL void cutt(int);
int main()
{
open("cell");
n = read(); m1 = read(); m2 = read();
cut(m1);
if (m1 == 1)
{
cout << "0" << endl;
return 0;
}
int ans = 999999999;
for (int i=1; i<=n; ++i)
{
a = read();
int temp = -1;
for (int j=1; j<=lend; ++j)
{
if (a%d[j].add)
{
temp = -1;
break;
}
int sum = 0;
while (!(a%d[j].add))
{
sum++;
a /= d[j].add;
}
sum = ceil(d[j].su*1.0/sum);
temp = max(temp, sum);
}
if (temp != -1)
ans = min(ans, temp);
}
if (ans == 999999999)
cout << "-1";
else cout << ans;
cout << endl;
return 0;
}
void cut(int x)
{
lend = 0;
for (int i=2; i*i<=x; ++i)
{
if (!(x%i))
{
lend++;
d[lend].add = i;
d[lend].su = 0;
while (!(x%i))
{
d[lend].su++;
x /= i;
}
d[lend].su *= m2;
}
}
if (x != 1)
{
d[++lend].add = x;
d[lend].su = m2;
}
}
int read()
{
int i = 1, j = 0;
char x = getchar();
while (x < '0' || '9' < x)
{
if ('-' == x) i = -1;
x = getchar();
}
while ('0' <= x && x <= '9')
{
j = j * 10 + x - '0';
x = getchar();
}
return i*j;
}这道题在考试的时候没有找到最优化的方法导致复杂度超高，然后re
导弹拦截
说是一道贪心题目，实际跟暴力枚举没有什么区别。现将最开始的情况直接全部分配给第一个系统并且要从大到小排序，在逐步枚举分配给第二个系统的截止位置。记录最小答案即可
#include<bits/stdc++.h>
#define open(s) freopen(s".in", "r", stdin); freopen(s".out", "w", stdout);
#define IL inline
#define ull unsigned long long
#define ll long long
using namespace std;
int ax, ay, bx, by;
int n, k = 1;
int xx[100010], yy[100010];
struct date
{
int x, y;
}a[100010];
IL int read();
IL int max(int, int);
IL bool cmp(date, date);
int main()
{
// open("missile");
ax = read(); ay = read(); bx = read(); by = read();
n = read();
for (int i=1; i<=n; ++i)
{
int x, y;
x = read(), y = read();
a[i].x = (ax - x) * (ax - x) + (ay - y) * (ay - y);
a[i].y = (bx - x) * (bx - x) + (by - y) * (by - y);
}
sort(a+1, a+1+n, cmp);
int i=2,r1=a[1].x,r2=0;
int minn = r1 + r2;
while (i <= n)
{
r1 = a[i].x;
r2 = max(r2, a[i-1].y);
minn = min(minn, r1+r2);
i++;
}
cout << minn << endl;
return 0;
}
bool cmp(date x, date y)
{
return x.x > y.x;
}
int max(int x, int y)
{
return x > y ? x : y;
}
int read()
{
int i = 1, j = 0;
char x = getchar();
while (x < '0' || '9' < x)
{
if ('-' == x) i = -1;
x = getchar();
}
while ('0' <= x && x <= '9')
{
j = j * 10 + x - '0';
x = getchar();
}
return i*j;
}这个题目直接贪心好像有一定的问题存在。所以必须保证平方和最小，而不是单个系统的最值问题
三国游戏
因为机器总是被动防守，所以总是选择不到大的值，而最大值的另外的搭配总是先被机器选走。所以人只能拿到次大，就直接寻找每一组搭配的次大值就可以了
#include<bits/stdc++.h>
#define open(s) freopen(s".in", "r", stdin); freopen(s".out", "w", stdout);
#define IL inline
#define ull unsigned long long
#define ll long long
using namespace std;
int n;
int a[510][510];
IL int read();
IL bool cmp(int x, int y)
{
return x > y;
}
int main()
{
//open("sanguo");
int maxn = -1, addj;
n = read();
for (int i=1; i<=n; ++i)
for (int j=i+1; j<=n; ++j)
a[i][j] = a[j][i] = read();
int ans = 0;
for (int i=1; i<=n; ++i)
{
sort(a[i]+1, a[i]+n+1, cmp);
maxn = max(maxn, a[i][2]);
}
cout << "1" << endl;
cout << maxn << endl;
return 0;
}
int read()
{
int i = 1, j = 0;
char x = getchar();
while (x < '0' || '9' < x)
{
if ('-' == x) i = -1;
x = getchar();
}
while ('0' <= x && x <= '9')
{
j = j * 10 + x - '0';
x = getchar();
}
return i*j;
}