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Farmer John is an astounding accounting wizard and has realized he might run out of money to run the farm. He has already calculated and recorded the exact amount of money (1 ≤ moneyi ≤ 10,000) that he will need to spend each day over the next N (1 ≤ N ≤ 100,000) days.
FJ wants to create a budget for a sequential set of exactly M (1 ≤ M ≤ N) fiscal periods called "fajomonths". Each of these fajomonths contains a set of 1 or more consecutive days. Every day is contained in exactly one fajomonth.
FJ‘s goal is to arrange the fajomonths so as to minimize the expenses of the fajomonth with the highest spending and thus determine his monthly spending limit.
Input
Output
Sample Input
7 5 100 400 300 100 500 101 400
Sample Output
500
Hint
/*
题意：给你n个数据让你分成m个块，使得这m个块的sum里面的最大的最小
二分 l应该要是a[i]的最大值， WA了一发。。注意1e4 * 1e5 = 1e9 不超int
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAX = 1e6 + 10;
int n, m;
int a[MAX];
bool check(int x)
{
int sum = 0, b = 0;
int cout = 1;
for(int i = 1; i <= n ; i++){
if(sum + a[i]> x){
cout++;
sum = a[i];
}
else sum += a[i];
}
if(cout > m) return false;
return true;
}
int main()
{
while(~scanf("%d%d", &n, &m)){
int l = 0, r = 1e9 + 100;
for(int i = 1; i <= n ; i++){
scanf("%d", &a[i]);
l = max(l, a[i]);
}
int ans = 0;
while(l <= r){
int mid = (l + r) >> 1;
if(check(mid)){
ans = mid;
r = mid - 1;
}
else l = mid + 1;
}
printf("%d\n", ans);
}
return 0;
}
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原文地址：http://www.cnblogs.com/zero-begin/p/4676082.html
