UVA - 409 Excuses, Excuses!
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http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=19139
做字符串的题总是如此费力,就更加说明了自己编程能力很欠缺,总是不能让问题简化,而是更加复杂化,抓不住问题的核心,像学长说的,要善于总结,并且要用最精简的代码,任重道远啊!
这道题还是水题,先给定n个字符串,然后给定m行字符串,让你输出n个字符串出现次数最多的那一行字符串,注意不区分大小写,所以可以把每一行字符都预处理成小写字母,
关键点在于 这句话 A keyword ``occurs" in an excuse if and only if it exists in the string in contiguous form and is delimited by the beginning or end of the line or any non-alphabetic character or a space.
我开始一直没明白什么意思,自然导致不停的wrong,但是后来意识到,一个字符串在一行中出现前后都是不能有字母的,和平时习惯一样,注意这点就没难度了。
#include <iostream>#include <cstdio>#include <cmath>#include <vector>#include <cstring>#include <string>#include <algorithm>#include <string>#include <set>#include <functional>#include <numeric>#include <sstream>#include <stack>#include <map>#include <queue>#define CL(arr, val) memset(arr, val, sizeof(arr))#define ll long long#define inf 0x7f7f7f7f#define lc l,m,rt<<1#define rc m + 1,r,rt<<1|1#define pi acos(-1.0)#define L(x) (x) << 1#define R(x) (x) << 1 | 1#define MID(l, r) (l + r) >> 1#define Min(x, y) (x) < (y) ? (x) : (y)#define Max(x, y) (x) < (y) ? (y) : (x)#define E(x) (1 << (x))#define iabs(x) (x) < 0 ? -(x) : (x)#define OUT(x) printf("%I64d\n", x)#define lowbit(x) (x)&(-x)#define Read() freopen("a.txt", "r", stdin)#define Write() freopen("dout.txt", "w", stdout);#define N 100005using namespace std;int solve(char *s2,char *s1){ int i,k,ans=0; //s1是关键词,s2是当前要查找的行 int l1=strlen(s1),l2=strlen(s2); for(i=0;i<l2;i++) { if(s2[i]==s1[0]&&(!(s2[i-1]>='a'&&s2[i-1]<='z'))) //保证前面不是字母 { for(k=1;k<l1;k++) //然后依次判断后面的字符 if(s1[k]!=s2[i+k]) break; if(!(s2[i+l1]>='a'&&s2[i+l1]<='z')&&k==l1) ans++; //保证后面不是字母 } } return ans;}int main(){ //Read(); int n,m,i,j,k=1; int f[25]; char s1[25][100],s2[25][100],s3[25][100]; while(~scanf("%d%d",&n,&m)) { getchar(); CL(f,0); CL(s3,0); for(i=0;i<n;i++) { gets(s1[i]); } for(i=0;i<m;i++) { gets(s2[i]); strcpy(s3[i],s2[i]); //把每一行都预先保存下来,输出的时候要用 int l=strlen(s2[i]); for(j=0;j<l;j++) if(isupper(s2[i][j])) s2[i][j]=tolower(s2[i][j]); //转换成小写字母 } int ans; for(i=0;i<m;i++) { //两重循环枚举字符串在一行中出现的字数 for(j=0;j<n;j++) { ans=0; ans=solve(s2[i],s1[j]); f[i]+=ans; //ans返回出现次数,f[i]保存关键词在第i行总共出现的次数 } } ans=0; for(i=0;i<m;i++) { if(f[i]>ans) ans=f[i]; //找出出现次数最多 的 } printf("Excuse Set #%d\n",k++); for(i=0;i<m;i++) { if(f[i]==ans) printf("%s\n",s3[i]); } printf("\n"); } return 0;}
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