HDU 4287 (13.08.17)
2 : a, b, c 3 : d, e, f 4 : g, h, i 5 : j, k, l 6 : m, n, o
7 : p, q, r, s 8 : t, u, v 9 : w, x, y, z
When we want to input the word “wing”, we press the button 9, 4, 6, 4, then the input method will choose from an embedded dictionary, all words matching the input number sequence, such as “wing”, “whoi”, “zhog”. Here comes our question, given a dictionary, how many words in it match some input number sequences?
Two integer N (1 <= N <= 5000), M (1 <= M <= 5000), indicating the number of input number sequences and the number of words in the dictionary, respectively. Then comes N lines, each line contains a number sequence, consisting of no more than 6 digits. Then comes M lines, each line contains a letter string, consisting of no more than 6 lower letters. It is guaranteed that there are neither duplicated number sequences nor duplicated words.
3 5
46
64448
74
go
in
night
might
gn
2
0
题意与思路:
输入N个按键序列, 再输入M个单词, 要求我们对M个单词进行检索, 求出其按键序列
然后统计, N个按键序列每个分别出现了几次
错误点: 由于不注意严谨, 以为少写一个break没事, 导致超时~
我有两份代码, 一份是错误的, 还没找出错误来, 一份已AC;
AC代码:
#include<stdio.h>
#include<string.h> int t[26] = {2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 9, 9, 9, 9}; int T, N, M; int ans[5555];
char num[5555][10];
char word[5555][10]; void getans(char *p) {
int i, len;
int pos = 0;
char t_num[10];
len = strlen(p);
for(i = 0; i < len; i++)
t_num[pos++] = t[p[i]-'a'] + '0';
t_num[pos] = '\0';
for(i = 0; i < N; i++)
if(strcmp(t_num, num[i]) == 0) {
ans[i]++;
break;
}
} int main() {
scanf("%d", &T);
while(T--) {
int i;
scanf("%d %d", &N, &M); getchar(); memset(ans, 0, sizeof(ans));
for(i = 0; i < N; i++)
gets(num[i]);
for(i = 0; i < M; i++)
gets(word[i]); for(i = 0; i < M; i++)
getans(word[i]); for(i = 0; i < N; i++)
printf("%d\n", ans[i]);
}
return 0;
}
wrong代码:
#include<stdio.h>
#include<string.h> int T, N, M; char num[5555][10];
char word[5555][10];
int ans[5555]; char t[10][10]; void getans(char *p) {
int i, j, k, pos = 0;
char t_num[10];
int len1, len2;
len1 = strlen(p);
for(i = 0; i < len1; i++) {
for(j = 2; j < 10; j++) {
len2 = strlen(t[j]);
for(k = 0; k < len2; k++) {
if(t[j][k] == p[i])
t_num[pos++] = j;
}
}
} t_num[pos] = '\0';
for(i = 0; i < N; i++) {
if(strcmp(t_num, num[i]) == 0)
ans[i]++;
}
} int main() {
scanf("%d", &T); while(T--) {
int i;
scanf("%d%d", &N, &M); memset(ans, 0, sizeof(ans));
strcpy(t[2], "abc");
strcpy(t[3], "def");
strcpy(t[4], "ghi");
strcpy(t[5], "jkl");
strcpy(t[6], "mno");
strcpy(t[7], "pqrs");
strcpy(t[8], "tuv");
strcpy(t[9], "wxyz"); for(i = 0; i < N; i++)
gets(num[i]);
for(i = 0; i < M; i++)
gets(word[i]); for(i = 0; i < M; i++)
getans(word[i]); for(i = 0; i < N; i++)
printf("%d\n", ans[i]);
}
return 0;
}
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