zoukankan      html  css  js  c++  java
  • CodeForces Round #521 (Div.3) A. Frog Jumping

    http://codeforces.com/contest/1077/problem/A

    A frog is currently at the point 00 on a coordinate axis OxOx. It jumps by the following algorithm: the first jump is aa units to the right, the second jump is bb units to the left, the third jump is aa units to the right, the fourth jump is bb units to the left, and so on.

    Formally:

    • if the frog has jumped an even number of times (before the current jump), it jumps from its current position xx to position x+ax+a;
    • otherwise it jumps from its current position xx to position xbx−b.

    Your task is to calculate the position of the frog after kk jumps.

    But... One more thing. You are watching tt different frogs so you have to answer tt independent queries.

    Input

    The first line of the input contains one integer tt (1t10001≤t≤1000) — the number of queries.

    Each of the next tt lines contain queries (one query per line).

    The query is described as three space-separated integers a,b,ka,b,k (1a,b,k1091≤a,b,k≤109) — the lengths of two types of jumps and the number of jumps, respectively.

    Output

    Print tt integers. The ii-th integer should be the answer for the ii-th query.

    Example
    input
    Copy
    6
    5 2 3
    100 1 4
    1 10 5
    1000000000 1 6
    1 1 1000000000
    1 1 999999999
    
    output
    Copy
    8
    198
    -17
    2999999997
    0
    1

    代码:

    #include <bits/stdc++.h>
    using namespace std;
    
    int T;
    
    int main() {
        scanf("%d", &T);
        while(T --) {
           long long a, b, k;
           cin >>a >> b >> k;
           long long ans = 0;
           if(a == b) {
                if(k % 2 == 0) ans = 0;
                else ans = a;
           } else {
                if(k % 2 == 0)
                    ans = (k / 2) * (a - b);
                else
                    ans = (k / 2) * (a - b) + a;
           }
           cout << ans << endl;
        }
        return 0;
    }
    

      

  • 相关阅读:
    简单 dp 题选做
    UVa11327
    Codeforces Round #641 (div.2) 题解
    新博客
    数位dp的学习
    stl粗略用法
    cf437C The Child and Toy
    poj1995 Raising Modulo Numbers
    Tarjan的学习
    最短路模板
  • 原文地址:https://www.cnblogs.com/zlrrrr/p/9990591.html
Copyright © 2011-2022 走看看