Monthly Expense
| Time Limit: 2000MS |
|
Memory Limit: 65536K |
| Total Submissions: 8261 |
|
Accepted: 3399 |
Description
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
Line 1: Two space-separated integers: N and M
Lines 2..N+1: Line i+1 contains the number of dollars Farmer John spends on the ith day
Output
Line 1: The smallest possible monthly limit Farmer John can afford to live with.
Sample Input
7 5
100
400
300
100
500
101
400
Sample Output
500
Hint
If Farmer John schedules the months so that the first two days are a month, the third and fourth are a month, and the last three are their own months, he spends at most $500 in any month. Any other method of scheduling gives a larger minimum monthly limit.
題意很簡單，算法也很簡單，以前這種題目絕對做不出來，可能沒做過這種類型的吧
現(xiàn)在看貌似很簡單，
做法就是二分枚舉答案+貪心驗證
二分的下界取最大的數(shù)
上界取所有數(shù)的和即可
唔，昨晚上剛看到這題的時候，想著dp應該是可以的
不過測試數(shù)據(jù)是10w，又不行了，然后想著可以轉(zhuǎn)化成圖論的模型，不過點太多，還是不行
然后搜索，dp都不行了，搜索也白搭，然后想二分，然后算了下只有10^9 ，頂多30多次 ，貌似可以
…………
#include<stdio.h>
#include<string.h>
#include<math.h>
#define maxn 100005
int n,m,a[maxn];
int lower,upper;
int max(int a,int b)
{
return a>b?a:b;
}
bool yanz(int x)
{
int num,i,tmp;
num=0;
i=1;
while(i<=n)
{
tmp=0;
while(tmp+a[i]<=x&&i<=n)
{
tmp=tmp+a[i];
i++;
}
num++;
}
if(num>m) return 0;
return 1;
}
int main()
{
int i;
while(scanf("%d%d",&n,&m)!=EOF)
{
lower=-1;
upper=0;
for(i=1; i<=n; i++)
{
scanf("%d",&a[i]);
lower=max(a[i],lower);
upper=upper+a[i];
}
//printf("%d %d\n",upper,lower);
int left,right;
left=lower;
right=upper;
while(left<right)
{
int mid=(left+right)/2;
if(yanz(mid)) right=mid; //printf("%d ok\n",right);}
else left=mid+1;
}
printf("%d\n",left);
}
return 0;
}
wa了兩遍，發(fā)現(xiàn)最開始的時候tmp沒初始化