poj-2803 Defining Moment
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 1660 | Accepted: 760 |
Description
For this problem, you'll be writing the prefix/suffix processing portion of the program.
Valid prefixes and their meanings are:
anti<word> | against <word> |
post<word> | after <word> |
pre<word> | before <word> |
re<word> | <word> again |
un<word> | not <word> |
Valid suffixes and their meanings are:
<word>er | one who <word>s |
<word>ing | to actively <word> |
<word>ize | change into <word> |
<word>s | multiple instances of <word> |
<word>tion | the process of <word>ing |
Note that suffixes are tied more tightly to their root word and
should therefore be expanded last. For example, the word ``vaporize"
would be expanded through the following steps:
unvaporize
not vaporize
not change into vapor
Of course, the definitions are not exactly right, but how much polish does the professor expect for a single homework grade?
Input
this problem will begin with a line containing a single integer n
indicating the number of words to define. Each of the following n lines
will contain a single word. You need to expand at most one prefix and
one suffix, and each word is guaranteed to have a non-empty root (i.e.,
if the prefix and/or suffix are removed, a non-empty string will
remain). Each word will be composed of no more than 100 printable
characters.
Output
Sample Input
6
vaporize
prewar
recooking
root
repopularize
uninforming
Sample Output
change into vapor
before war
to actively cook again
root
change into popular again
not to actively inform
Source
poj-2803
Caution_X
20191004
tags:
模拟
description modelling:
处理至多一个前缀和后缀
major steps to solve it:
1.先处理前缀后将字符串前缀删去
2.处理后缀
#include<cstdio>
using namespace std;
int n,k;
int a[][];
int main()
{
//freopen("input.txt","r",stdin);
while(~scanf("%d%d",&n,&k)&&n&&k)
{
for(int i=;i<k;i++) {
for(int j=;j<n;j++) {
scanf("%d",&a[i][j]);
}
}
bool ok=false;
for(int i=;i<n;i++) {
int sum=;
for(int j=;j<k;j++) {
sum+=a[j][i];
}
if(sum==k) {
ok=true;
break;
}
}
if(ok) printf("yes\n");
else printf("no\n");
}
return ;
}
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