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本文链接：https://blog.youkuaiyun.com/eunyeon/article/details/43167679

本文讨论了如何通过修改字符数组长度解决POJ平台上的编程挑战题，涉及字符串匹配、排序算法及字符映射的应用。

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一开始char数组长度30时WA，改成200就通过了。

1128K 579MS

#include <iostream><span style="font-family:Arial, Helvetica, sans-serif;"><span style="white-space: normal;">  </span></span>
#include <algorithm>
#include <string.h>
using namespace std;
struct phone_number{
	char num[10];
}pn[100005];
char map[26] = {'2','2','2','3','3','3','4','4','4','5','5','5','6','6','6','7',0,'7','7','8','8','8','9','9','9',0};
char s[200];
int amt,flag;

int cmp(struct phone_number &pn1,struct phone_number &pn2){
	return strcmp(pn1.num,pn2.num)<0;
}

int main(){
	int i,j,k;
	scanf("%d",&amt);
	for(i=0;i<amt;i++){
		scanf("%s",s);
		k=0;
		for(j=0;j<strlen(s);j++){
			if(s[j]=='-'||s[j]=='Q'||s[j]=='Z') continue;
			if(s[j]>='A'&&s[j]<='Z')
				pn[i].num[k++]=map[s[j]-'A'];
			else pn[i].num[k++]=s[j];
			if(k==3) pn[i].num[k++]='-';
		}
	}
	sort(pn,pn+amt,cmp);
	k=1;
	for(i=0;i<amt;i++){
		if(strcmp(pn[i].num,pn[i+1].num)==0){
			k++;
		}
		else{
			if(k>1){
				printf("%s %d\n",pn[i].num,k);
				flag=1;
			}
			k=1;
		}
	}
	if(flag==0) printf("No duplicates.\n");
	return 0;
}