Intel Code Challenge Final Round (Div. 1 + Div. 2, Combined) D题
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题意:
给出一串字符串,每M个字符中必须取一个字符,问字典序最小为多少。
思路:
依次从A B- Z 中取出字符。
如果有一个M段 没有当前取的字符CH ,那么就把这个字符串中所有的CH输出,作为最小,并且标记。
如果都有CH,那么在最后按位置把CH最小的个数输出
#include<cstdio>#include<cstring>#include<iostream>#include<vector>#include<algorithm>#include<cmath>#include<map>#include<set>#include<queue>using namespace std; int m;char st[100010];bool b[100010];int main(){ scanf("%d",&m); scanf("%s",st); for(char ch='a'; ch<='z'; ch++) { int sum=0; for(int i=0; st[i]!='\0'; i++) { if (st[i]==ch||b[i]) sum=0; else sum++; if (sum==m) break; } if (sum==m) { for(int i=0; st[i]!='\0'; i++) if (st[i]==ch) { printf("%c",ch); b[i]=true; } continue; } sum=0; int l=-1; for(int i=0; st[i]!='\0'; i++) { if (b[i]) sum=0; else sum++; if (st[i]==ch) l=i; if (sum==m) { sum=i-l; printf("%c",ch); } } break; } cout<<endl; return 0;}
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