bzoj1676[Usaco2005 Feb]Feed Accounting 饲料计算
Description
Farmer John is trying to figure out when his last shipment of feed arrived. Starting with an empty grain bin, he ordered and received F1 (1 <= F1 <= 1,000,000) kilograms
of feed. Regrettably, he is not certain exactly when the feed arrived. Of the F1 kilograms, F2 (1 <= F2 <= F1) kilograms of feed remain on day D (1 <= D <= 2,000). He must determine the most recent day that his shipment could have arrived. Each of his C (1
<= C <= 100) cows eats exactly 1 kilogram of feed each day. For various reasons, cows arrive on a certain day and depart on another, so two days might have very different feed consumption. The input data tells which days each cow was present. Every cow ate
feed from Farmer John's bin on the day she arrived and also on the day she left. Given that today is day D, determine the minimum number of days that must have passed since his last shipment. The cows have already eaten today, and the shipment arrived before
the cows had eaten.
Input
* Line 1: Four space-separated integers: C, F1, F2, and D * Lines 2..C+1: Line i+1 contains two space-separated integers describing the presence of a cow. The first integer tells the first day the cow
was on the farm; the second tells the final day of the cow's presence. Each day is in the range 1..2,000.
Output
The last day that the shipment might have arrived, an integer that will always be positive.
Sample Input
1 9
5 8
8 12
INPUT DETAILS:
The shipment was 14 kilograms of feed, and Farmer John has 4 kilograms
left. He had three cows that ate feed for some amount of time in
the last 10 days.
Sample Output
上一次运来了14千克饲料,现在饲料还剩下4千克.最近10天里.有3头牛来吃过饲料.
约翰在第6天收到14千克饲料,当天吃掉2千克,第7天吃掉2千克,第8天吃掉3千克,第9天吃掉2千克,第10天吃掉1千克,正好还剩4千克
倒着搞差分序列……f2每次更新……到f2>=f1的时候就输出……没了
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int c,f1,f2,d,r,sum;
int a[2010];
int main()
{
c=read();
f1=read();
f2=read();
d=read();
while (c--)
{
int x=read(),y=read();
a[x-1]--;a[y]++;
r=max(r,y);
}
for (int i=r;i>0;i--)
{
sum+=a[i];
if(i<=d)f2+=sum;
if (f2>=f1)
{
printf("%d",i);
return 0;
}
}
}
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