codeforces790A - Bear and Different Names
2017-11-17 20:46
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题面在这里
题目大意:
就是说有一个长度为n的名字序列,然后告诉你每连续m个里有没有重复的,有重复的就是NO,没有重复就是YES。让你构造一个合法的名字序列。
一个简单的贪心思想,前m-1个赋值为1~m-1,每次有重复的就让他等于这m个的第一个,这样就不会对后面有影响。如果没有重复的就再新建一个。
/*************************************************************
Problem: codeforces 790A - Bear and Different Names
User: fengyuan
Language: C++
Result: Accepted
Time: 15 ms
Memory: 0 KB
Submit_Time: 2017-11-17 20:40:50
*************************************************************/
#include<cstdio>
#include<cstring>
using namespace std;
const int N = 55;
char s[5];
int ans
, cnt = 0, n, m;
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i < m; i ++){
ans[i] = i;
cnt ++;
}
for (int i = 1; i <= n-m+1; i ++){
scanf("%s", s);
if (s[0] == 'N'){
ans[i+m-1] = ans[i];
} else {
ans[i+m-1] = ++ cnt;
}
}
for (int i = 1; i <= n; i ++){
if (ans[i] <= 26){
putchar(ans[i]+'A'-1);
putchar(' ');
} else {
putchar(ans[i]+'A'-27);
putchar(ans[i]+'a'-27);
putchar(' ');
}
}
return 0;
}
题目大意:
就是说有一个长度为n的名字序列,然后告诉你每连续m个里有没有重复的,有重复的就是NO,没有重复就是YES。让你构造一个合法的名字序列。
一个简单的贪心思想,前m-1个赋值为1~m-1,每次有重复的就让他等于这m个的第一个,这样就不会对后面有影响。如果没有重复的就再新建一个。
/*************************************************************
Problem: codeforces 790A - Bear and Different Names
User: fengyuan
Language: C++
Result: Accepted
Time: 15 ms
Memory: 0 KB
Submit_Time: 2017-11-17 20:40:50
*************************************************************/
#include<cstdio>
#include<cstring>
using namespace std;
const int N = 55;
char s[5];
int ans
, cnt = 0, n, m;
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i < m; i ++){
ans[i] = i;
cnt ++;
}
for (int i = 1; i <= n-m+1; i ++){
scanf("%s", s);
if (s[0] == 'N'){
ans[i+m-1] = ans[i];
} else {
ans[i+m-1] = ++ cnt;
}
}
for (int i = 1; i <= n; i ++){
if (ans[i] <= 26){
putchar(ans[i]+'A'-1);
putchar(' ');
} else {
putchar(ans[i]+'A'-27);
putchar(ans[i]+'a'-27);
putchar(' ');
}
}
return 0;
}
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