B. Covered Path

Time Limit: 1 Sec Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/534/problem/B

Description

The on-board computer on Polycarp's car measured that the car speed at the beginning of some section of the path equals v₁ meters per second, and in the end it is v₂ meters per second. We know that this section of the route took exactly t seconds to pass.

Assuming that at each of the seconds the speed is constant, and between seconds the speed can change at most by d meters per second in absolute value (i.e., the difference in the speed of any two adjacent seconds does not exceed d in absolute value), find the maximum possible length of the path section in meters.

Input

The first line contains two integers v₁ and v₂ (1 ≤ v₁, v₂ ≤ 100) — the speeds in meters per second at the beginning of the segment and at the end of the segment, respectively.

The second line contains two integers t (2 ≤ t ≤ 100) — the time when the car moves along the segment in seconds, d (0 ≤ d ≤ 10) — the maximum value of the speed change between adjacent seconds.

It is guaranteed that there is a way to complete the segment so that:

the speed in the first second equals v₁,

the speed in the last second equals v₂,

the absolute value of difference of speeds between any two adjacent seconds doesn't exceed d.

Output

Print the maximum possible length of the path segment in meters.

Sample Input

           input1 

           5 6 
4 2 

           input2 

           10 10 
10 0 

Sample Output

26

100

HINT

题意

啊，给你初速度和末速度，给你运动时间，和最大加速度

当然，每一秒的速度都是整数，然后问你在这种情况下，路程最大是多少

题解：

啊，暴力枚举每秒钟加多少就好啦，肯定是越早加速越好啦

代码：

//qscqesze#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
               #include 
               
                #include 
                
                 #include 
                 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 200001#define mod 10007#define eps 1e-9//const int inf=0x7fffffff; //无限大const int inf=0x3f3f3f3f;/*inline ll 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 buf[10];inline void write(int i) { int p = 0;if(i == 0) p++; else while(i) {buf[p++] = i % 10;i /= 10;} for(int j = p-1; j >=0; j--) putchar('0' + buf[j]); printf("\n");}*///**************************************************************************************int main(){ int v1,v2; cin>>v1>>v2; int d,t; cin>>t>>d; int ans=v1; for(int i=2;i<=t;i++) { for(int j=d;j>=-d;j--) { if(v1+j-(t-i)*d<=v2) { v1=v1+j; break; } } ans+=v1; //cout<
                  
                   <

转载地址：http://wivka.baihongyu.com/

你可能感兴趣的文章