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- ;
int Num;
//const int inf=0x7fffffff;
const ll inf=;
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
//************************************************************************************* int a[maxn];
int main()
{
int t=read();
for(int cas=;cas<=t;cas++)
{
int n=read(),A=read(),B=read(),l=read();
memset(a,,sizeof(a));
for(int i=;i<=n;i++)
{
int L=read(),R=read();
for(int j=L;j<R;j++)
{
a[j]=;
}
}
int temp = ;
int ans = ;
for(int i=;i<l;i++)
{
if(a[i]==)
{
if(temp<A)
{
ans+=A-temp;
temp=;
}
else temp=temp-A;
}
else
temp+=B;
}
printf("Case #%d: %d\n",cas,ans);
}
}
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