2020.07.28【NOIP提高组】模拟
考试时状态不好,暴力不想打
结束前勉勉强强骗点分
已经不想说什么了······
(T1) 复制&粘贴2
逆推答案,枚举 (k),分类讨论
(T2) 愉快的logo设计
暴力计算贡献改为前缀和,把原串复制一片,把目标串分成连续的多块,记录开头结尾,根据前缀和算成功匹配的个数,取最大的即可
(T3) Flower 特难的题,见我另一篇,这里不展开叙述
(T4) 有趣的有趣的家庭菜园 还可以,见我另一篇,这里不展开叙述
前两题代码:
(Code1)
#include<cstdio>
#include<cstring>
using namespace std;
const int N = 2e5 + 5;
int k , m , n , x[N] , y[N] , z[N];
char s[N];
int main()
{
freopen("copypaste.in" , "r" , stdin);
freopen("copypaste.out" , "w" , stdout);
scanf("%d%d%s%d" , &k , &m , s , &n);
for(register int i = 1; i <= n; i++) scanf("%d%d%d" , &x[i] , &y[i] , &z[i]);
int len = strlen(s);
for(register int i = len; i; i--) s[i] = s[i - 1];
for(register int i = 1; i <= k; i++)
{
int ans = i;
for(register int j = n; j; j--)
{
if (z[j] + y[j] - x[j] < ans) ans = ans - (y[j] - x[j]);
else if (ans > z[j]) ans = ans - z[j] + x[j];
}
printf("%c" , s[ans]);
}
}
(Code2)
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
const int N = 5000005;
int k , ans = N , tot = 0 , num[5][N];
char s[N] , str[13][N];
struct node{
int l , r , s;
}fl[N];
int main()
{
freopen("logo.in" , "r" , stdin);
freopen("logo.out" , "w" , stdout);
scanf("%d%s" , &k , s);
int p = 1;
str[0][0] = '0';
for(register int i = 1; i <= k; i++)
{
for(register int j = 1; j <= p; j++) str[i][j - 1] = 'J' , str[i][j + p - 1] = 'O' , str[i][j + p * 2 - 1] = 'I';
for(register int j = 3 * p; j < 4 * p; j++) str[i][j] = str[i - 1][j - 3 * p];
p *= 4;
}
for(register int i = 0; i < p - 1; i++)
{
int j = i;
fl[++tot].l = i;
fl[tot].s = str[k][i] == 'J' ? 1 : (str[k][i] == 'O' ? 2 : 3);
while (str[k][j] == str[k][j + 1]) j++;
fl[tot].r = j;
i = j;
}
p = strlen(s);
for(register int i = p; i <= p * 2 - 1; i++) s[i] = s[i - p];
for(register int i = 0; i <= p * 2 - 1; i++)
{
int x = s[i] == 'J' ? 1 : (s[i] == 'O' ? 2 : 3);
if (i != 0)
{
num[1][i] = num[1][i - 1];
num[2][i] = num[2][i - 1];
num[3][i] = num[3][i - 1];
}
num[x][i]++;
}
for(register int i = 0; i < p; i++)
{
int x = 0;
for(register int j = 1; j <= tot; j++)
x += fl[j].l + i != 0 ? num[fl[j].s][fl[j].r + i] - num[fl[j].s][fl[j].l + i - 1] : num[fl[j].s][fl[j].r + i];
ans = min(ans , p - x);
}
printf("%d" , ans - 1);
}