UVA 12482 Short Story Competition
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Maratona de Programação da SBC – ACM ICPC – 2012
Problem A
Short Story Competition
Machado wants to be a writer. He has written many short stories, book reviews, reports of trips
he made, and a little romance. Now Machado wants to take part in a short story competition, which
has very strict rules about the submission format.
The rules of the competition limit the number of total pages, and specify the maximum number
of characters per line, and the maximum number of lines per page. Additionally, each word must be
written integrally in one line (ie, a word cannot be separated in two lines). Machado wants to write a
story with as many words as possible within the rules of the contest, and needs your help.
Given the maximum number of characters per line, the maximum number of lines per page, and
the words of the short story that Machado is writing, he wants to know the minimum number of pages
that his short story will ocuppy, considering the rules of the contest.
Input
The first line of a test case contains three integers N, L and C, which indicate, respectively, the
number of words of the short story, the maximum number of lines per page and the maximum number
of characters per line. Machado’s short story is innovative and contains no characters besides upper
and lower case letters and blanks. The second line contains Machado’s short story, consisting of N
words separated by exactly one blank space.
Output
For each test case your program must output a single line containing a single integer indicating the
minimum number of pages the short story will ocuppy, considering the contest rules.
Restrictions
• 2 ≤ N ≤ 1000
• 1 ≤ L ≤ 30
• 1 ≤ C ≤ 70
• 1 ≤ length of each word ≤ C
Example
Sample Input
14 4 20
Se mana Piedade tem casado com Quincas Borba apenas me daria uma esperanca colateral
16 3 30
No dia seguinte entrou a dizer de mim nomes feios e acabou alcunhando me Dom Casmurro
5 2 2
a de i de o
5 2 2
a e i o u
Sample output
2
1
3
3
题意:给你一篇短故事,求至少需要几页纸才能容纳这篇段故事,已知短故事有N个单词,每页能容纳L行,每行能容纳C个字符,这里要注意每个单词后有一个空格,一个单词只能在同一行。
思路:数据较小,直接定义二维数组算出每个单词的长度,然后一个个找下去即可。
#include<stdio.h>#include<string.h>char a[1111][77];int b[1111];int main(){int n,l,c,sum,flag,i;while(scanf("%d %d %d",&n,&l,&c)!=EOF){for(i=0;i<n;i++)scanf("%s",a[i]);for(i=0;i<n;i++)b[i]=strlen(a[i]);sum=0;flag=0;for(i=0;i<n;i++){if(flag+b[i]<=c){flag+=b[i];flag++;}else{sum++;flag=0;i--;}}if(flag!=0)sum++;if(sum%l==0)printf("%d\n",sum/l);elseprintf("%d\n",sum/l+1);}return 0;}
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