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 4Sample Output
Case #1: 0题意：走一段有沼泽有平路的路，走沼泽地时每米消耗a点体力，走平路时每秒恢复b点体力，求最开始需要最少的体力。
可以直接模拟这个状态就可以做出来，也有的大佬用二分做，我当时是直接进行的模拟。
简单模拟：
#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
using namespace std;
typedef long long ll;
struct ss{
int l;
int r;
}s[200];
int main()
{
int n,a,b,l,t;
cin>>t;
int k=0;
while(t--){
k++;
cin>>n>>a>>b>>l;
for(int i=0;i<n;i++){
scanf("%d %d",&s[i].l,&s[i].r);
}
int ans=0;
int temp=0;
int c=0;
for(int i=0;i<n;i++){
c+=b*(s[i].l-temp);
if(c<(s[i].r-s[i].l)*a){
ans+=((s[i].r-s[i].l)*a-c);
c=0;
}
else{
c-=(s[i].r-s[i].l)*a;
}
temp=s[i].r;
}
printf("Case #%d: %d\n",k,ans);
}
return 0;
}
二分：
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
const int maxn=110;
struct Node{
int l,r;
}A[maxn];
int a,b;
bool judge(int mid,int n){
int ans=mid,l=0;
for(int i=0;i<n;++i){
ans+=b*(A[i].l-l);
ans-=a*(A[i].r-A[i].l);
l=A[i].r;
if(ans<0)return false;
}
return true;
}
int main()
{
int t,i,j,k=1,l,n;
scanf("%d",&t);
while(t--){
scanf("%d%d%d%d",&n,&a,&b,&l);
int l=0,r=0;
for(i=0;i<n;++i){
scanf("%d%d",&A[i].l,&A[i].r);
r+=(A[i].r-A[i].l)*a;
}
printf("Case #%d: ",k++);
int ans=r;
while(l<=r){
int mid=(l+r)>>1;
if(judge(mid,n)){
ans=mid;
r=mid-1;
}
else {
l=mid+1;
}
}
printf("%d\n",ans);
}
return 0;
}