Warm up 16 Ragged Right

Ragged Right

Time Limit: 10000ms, Special Time Limit:25000ms, Memory Limit:65536KB

Total submit users: 38, Accepted users: 37

Problem 12786 : No special judgement

Problem description

又是一道感人的出入方式的傻逼题啊！

Word wrapping is the task of deciding how to break a paragraph of text into lines. For aesthetic reasons, we’d like all the lines except the last one to be about the same length. For example, we would say the text on the left looks less ragged than the text on the right: This is a This paragraph is a paragraph of text. of text. Your job is to compute a raggedness value for an arbitrary paragraph of text. We’ll measure raggedness in a way similar to the T EX typesetting system. Let n be the length, measured in characters, of the longest line of the paragraph.If some other line contains only m characters, then we’ll charge a penalty score of (n − m)^2 for that line. The raggedness will be the sum of the penalty scores for every line except the last one.

Input

Input consists of a single paragraph of text containing at most 100 lines. Each line of the paragraph contains a sequence of between 1 and 80 characters (letters, punctuation characters, decimal digits and spaces). No line starts or ends with spaces. The paragraph ends at end of file.

Output

Print out a single integer, the raggedness score for paragraph.

Sample Input

Sample Input 1
some blocks
of text line up
well on the right,
but
some don’t.

Sample Input 2
this line is short
this one is a bit longer
and this is the longest of all.

Sample Output

Sample Input 1
283

Sample Input 2
218

Problem Source

HNU Contest 

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#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <queue>
#include <vector>
#include <string>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define maxn 105
#define ll long long
#define mod 1000000007
#define INF 0x7fffffff
#define eps 1e-8
int n, m;
ll gcd(ll n, ll m){ return m ? gcd(m, n%m) : n; }
char s[maxn];
int a[maxn];
int main(){
    /*int t;
    scanf("%d", &t);
    while (t--){
    scanf("%I64d", &n);

    }*/
    int k = 0;
    int ma = -INF;
    while (1){
        if (gets(s) == NULL)break;
        n = strlen(s);
        a[k++] = n;
        if (ma < n)ma = n;
    }
        int sum = 0;
        for (int i = 0; i < k - 1; i++)
            sum += (ma - a[i]) * (ma-a[i]);
        printf("%d\n", sum);
    return 0;
}