题干：
The main city magazine offers its readers an opportunity to publish their ads. The format of the ad should be like this:
There are space-separated non-empty words of lowercase and uppercase Latin letters.
There are hyphen characters '-' in some words, their positions set word wrapping points. Word can include more than one hyphen.
It is guaranteed that there are no adjacent spaces and no adjacent hyphens. No hyphen is adjacent to space. There are no spaces and no hyphens before the first word and after the last word.
When the word is wrapped, the part of the word before hyphen and the hyphen itself stay on current line and the next part of the word is put on the next line. You can also put line break between two words, in that case the space stays on current line. Check notes for better understanding.
The ad can occupy no more that k lines and should have minimal width. The width of the ad is the maximal length of string (letters, spaces and hyphens are counted) in it.
You should write a program that will find minimal width of the ad.
Input
The first line contains number k (1 ≤ k ≤ 105).
The second line contains the text of the ad — non-empty space-separated words of lowercase and uppercase Latin letters and hyphens. Total length of the ad don't exceed 106 characters.
Output
Output minimal width of the ad.
Examples
Input
4 garage for sa-le
Output
7
Input
4 Edu-ca-tion-al Ro-unds are so fun
Output
10
Note
Here all spaces are replaced with dots.
In the first example one of possible results after all word wraps looks like this:
garage. for. sa- le
The second example:
Edu-ca- tion-al. Ro-unds. are.so.fun
解题报告：
以前做过一个类似的，直接二分枚举答案就可以了，注意fit中直接让cnt=1，表示，当前在凑第cnt个，这样设计cnt是为了表示最后一个点，需要多一个，所以干脆直接cnt=1好了，而且也有比较合理的解释。
AC代码：
#include<bits/stdc++.h>
using namespace std;
char s[1000005];
int ok[1000005];
int len,tot,k;
bool fit(int x) {
int cur = -1,cnt = 1;
for(int i = 1; i<=tot; i++) {
if(ok[i] - cur > x) {
i--;cnt++;cur = ok[i];
}
if(cnt > k) return 0;
}
return 1;
}
int main()
{
int maxx = 0;
cin>>k;getchar();
gets(s);
len = strlen(s);
ok[0] = -1;
for(int i = 0; i<len; i++) {
if(s[i] == ' ' ||s[i] == '-') {
ok[++tot] = i;
maxx = max(maxx,ok[tot] - ok[tot-1]);
}
}
ok[++tot] = len-1;
int l = maxx,r = len;
int mid = (l + r) / 2;
while(l < r) {
mid = (l + r ) / 2 ;
if(fit(mid)) r = mid;
else l = mid + 1;
}
printf("%d\n",l);
return 0 ;
} 总结：l=maxx开始，因为这样就避免了fit中需要先循环判断一次是否有一个串的长度也不符合的，r从len开始，其实从哪开始都可以，刚开始写的是len/k +1 不知道自己是怎么想的。。。肯定是错的反正。