Master Di plans to take his girlfriend for a travel by bike. Their journey, which can be seen as a line segment of length L, is a road of swamps and flats. In the swamp, it takes A point strengths per meter for Master Di to ride; In the flats, Master Di will regain B point strengths per meter when riding. Master Di wonders:In the beginning, he needs to prepare how much minimum strengths. (Except riding all the time,Master Di has no other choice)
Input
In the first line there is an integer t (1≤t≤501≤t≤50), indicating the number of test cases.
For each test case:
The first line contains four integers, n, A, B, L.
Next n lines, each line contains two integers: Li,RiLi,Ri, which represents the interval [Li,Ri][Li,Ri] is swamp.
1≤n≤100,1≤L≤105,1≤A≤10,1≤B≤10,1≤Li<Ri≤L1≤n≤100,1≤L≤105,1≤A≤10,1≤B≤10,1≤Li<Ri≤L.
Make sure intervals are not overlapped which means Ri<Li+1Ri<Li+1 for each i (1≤i<n1≤i<n).
Others are all flats except the swamps.
Output
For each text case:
Please output “Case #k: answer”(without quotes) one line, where k means the case number counting from 1, and the answer is his minimum strengths in the beginning.
Sample Input
1 2 2 2 5 1 2 3 4
Sample Output
Case #1: 0
cin 超时改成scanf AC
#include<bits/stdc++.h>
using namespace std;
int d[100005];
int main()
{
int n,a,b,l,p,q;
int t;
cin>>t;
for(int k=1;k<=t;k++)
{
scanf("%d%d%d%d",&n,&a,&b,&l);
for(int i=0;i<l;i++)
d[i]=b;
for(int i=0;i<n;i++)
{
scanf("%d%d",&p,&q);
for(int j=p;j<q;j++)
d[j]=-a;
}
int sum=0;
int ans=0;
for(int i=0;i<l;i++)
{
sum+=d[i];
if(sum<0)
{
ans+=abs(sum);
sum=0;
}
}
printf("Case #%d: %d\n",k,ans);
}
}