比赛入口
B Equal Rectangles
做法：当时一直在想是不是和lcm有关，然后各种模拟判断（去世…）其实只有头和尾才能凑成一对…头和尾相乘和其他对相等的概率是最大的。
代码：
#include<bits/stdc++.h>
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define lson rt<<1, l, mid
#define rson rt<<1|1, mid+1, r
using namespace std;
typedef long long LL;
typedef pair<LL, int> pli;
typedef pair<int, int> pii;
typedef pair<LL, LL> pll;
typedef pair<double, double> pdd;
typedef map<char, int> mci;
typedef map<string, int> msi;
template<class T>
void read(T &res) {
int f = 1; res = 0;
char c = getchar();
while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }
while(c >= '0' && c <= '9') { res = res * 10 + c - '0'; c = getchar(); }
res *= f;
}
int a[405];
int main() {
int t; read(t);
int n;
while(t--) {
read(n);
for(int i = 0; i < 4*n; ++i) read(a[i]);
sort(a, a+4*n);
int f = 1, mid = a[0] * a[4*n-1];
for(int i = 0; i < 4*n; i += 2) {
if(a[i] * a[4*n-i-1] != mid || a[i] != a[i+1]) {
f = 0; break;
}
}
if(f) puts("YES");
else puts("NO");
}
return 0;
}
D2 Remove the Substring (hard version)
做法：给出两个字符串，t字符串是s字符串删掉一些子串的结果，要求的是删掉子串中长度最长是多少。只需要记录最前面的位置和最后面的位置，再遍历一遍相减就好了…还是没想到呀…
代码：
#include<bits/stdc++.h>
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define lson rt<<1, l, mid
#define rson rt<<1|1, mid+1, r
using namespace std;
typedef long long LL;
typedef pair<LL, int> pli;
typedef pair<int, int> pii;
typedef pair<LL, LL> pll;
typedef pair<double, double> pdd;
typedef map<char, int> mci;
typedef map<string, int> msi;
template<class T>
void read(T &res) {
int f = 1; res = 0;
char c = getchar();
while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }
while(c >= '0' && c <= '9') { res = res * 10 + c - '0'; c = getchar(); }
res *= f;
}
const int N = 2e5+10;
char s[N], t[N];
int ls, lt, p1[N], p2[N], ans;
int main() {
scanf("%s%s", s+1, t+1);
ls = strlen(s+1); lt = strlen(t+1);
p1[0] = 0; p2[lt+1] = ls+1;
for(int i = 1, j = 0; i <= ls; ++i)
s[i] == t[j+1] ? p1[++j] = i : 0;
for(int i = ls, j = lt + 1; i; --i)
s[i] == t[j-1] ? p2[--j] = i : 0;
for(int i = 0; i <= lt; ++i)
ans = max(ans, p2[i+1] - p1[i] - 1);
printf("%d\n", ans);
return 0;
}
F2 Complete the Projects (hard version)
做法：最后一题应该是排序+dp了…
