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  • SPOJ AMR10F Cookies Piles

    AMR10F - Cookies Piles

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    The kids in my son's kindergarten made Christmas cookies with their teacher, and piled them up in columns.  They then arranged the columns so that the tops of the columns, going from shortest to tallest, were in a nice straight ramp.  The cookies were all of uniform size.  Given that there were A cookies in the shortest pile, that the difference in height between any two adjacent piles was D cookies, and that there were N piles, can you write a program to figure out how many cookies there were in total? 
     
    INPUT 
    The first line contains the number of test cases T. T lines follow, one corresponding to each test case, containing 3 integers : N, A and D. 
     
    OUTPUT
    Output T lines, each line containing the required answer for the corresponding test case. 
     
    CONSTRAINTS 
    T <= 100 
    1 <= N, A, D <=100 
     
    SAMPLE INPUT 

    1 1 1 
    3 5 6 
    2 1 2 
     
    SAMPLE OUTPUT 

    33 

     
    EXPLANATION 
    In the second test case the sequence is: 5, 11, 17 whose sum is 33.

    水.

    /*************************************************************************
        > File Name: code/2015summer/#4/F.cpp
        > Author: 111qqz
        > Email: rkz2013@126.com 
        > Created Time: 2015年07月29日 星期三 21时47分23秒
     ************************************************************************/
    
    #include<iostream>
    #include<iomanip>
    #include<cstdio>
    #include<algorithm>
    #include<cmath>
    #include<cstring>
    #include<string>
    #include<map>
    #include<set>
    #include<queue>
    #include<vector>
    #include<stack>
    #define y0 abc111qqz
    #define y1 hust111qqz
    #define yn hez111qqz
    #define j1 cute111qqz
    #define tm crazy111qqz
    #define lr dying111qqz
    using namespace std;
    #define REP(i, n) for (int i=0;i<int(n);++i)  
    typedef long long LL;
    typedef unsigned long long ULL;
    const int inf = 0x7fffffff;
    int main()
    {
        int T;
        int n,a,d;
        cin>>T;
        while (T--)
        {
        scanf("%d %d %d",&n,&a,&d);
        cout<<n*a+n*(n-1)/2*d<<endl;
        }
      
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/111qqz/p/4687548.html
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