HDU 5477 A Sweet Journey 水题

A Sweet Journey

Time Limit: 1 Sec  

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5477

Description

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≤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,Ri, which represents the interval [Li,Ri] is swamp.
1≤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+1 for each i (1≤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

HINT

题意

有一个人，走沼泽地会损失ai点能量，走正常的会得到bi点能量

然后问你一开始需要多少能量才行？

题解：

扫一遍就好了，签到题

代码：

//qscqesze
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <bitset>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 100006
#define mod 1000000007
#define eps 1e-9
#define e exp(1.0)
#define PI acos(-1)
const double EP  = 1E-10 ;
int Num;
//const int inf=0x7fffffff;
const ll inf=999999999;
inline ll read()
{
    ll 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 a[maxn];
int main()
{
    int t=read();
    for(int cas=1;cas<=t;cas++)
    {
        int n=read(),A=read(),B=read(),l=read();
        memset(a,0,sizeof(a));
        for(int i=1;i<=n;i++)
        {
            int L=read(),R=read();
            for(int j=L;j<R;j++)
            {
                a[j]=1;
            }
        }
        int temp = 0;
        int ans = 0;
        for(int i=0;i<l;i++)
        {
            if(a[i]==1)
            {
                if(temp<A)
                {
                    ans+=A-temp;
                    temp=0;
                }
                else temp=temp-A;
            }
            else
                temp+=B;
        }
        printf("Case #%d: %d
",cas,ans);
    }
}