ACM刷题之ZOJ————Hard to Play

Hard to Play

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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MightyHorse is playing a music game called osu!.

After playing for several months, MightyHorse discovered the way of calculating score in osu!:

1. While playing osu!, player need to click some circles following the rhythm. Each time a player clicks, it will have three different points: 300, 100 and 50, deciding by how clicking timing fits the music.

2. Calculating the score is quite simple. Each time player clicks and gets P points, the total score will add P, which should be calculated according to following formula:

P = Point * (Combo * 2 + 1)

Here Point is the point the player gets (300, 100 or 50) and Combo is the number of consecutive circles the player gets points previously - That means if the player doesn't miss any circle and clicks the i^th circle, Combo should be i - 1.

Recently MightyHorse meets a high-end osu! player. After watching his replay, MightyHorse finds that the game is very hard to play. But he is more interested in another problem: What's the maximum and minimum total score a player can get if he only knows the number of 300, 100 and 50 points the player gets in one play?

As the high-end player plays so well, we can assume that he won't miss any circle while playing osu!; Thus he can get at least 50 point for a circle.

Input

There are multiple test cases.

The first line of input is an integer T (1 ≤ T ≤ 100), indicating the number of test cases.

For each test case, there is only one line contains three integers: A (0 ≤ A ≤ 500) - the number of 300 point he gets, B (0 ≤ B ≤ 500) - the number of 100 point he gets and C (0 ≤ C ≤ 500) - the number of 50 point he gets.

Output

For each test case, output a line contains two integers, describing the minimum and maximum total score the player can get.

Sample Input

12 1 1 

Sample Output

2050 3950

水题

多个for暴力即可

下面是ac代码

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>#include<algorithm>#include<map>#include<set>#include<queue>#include<string>#include<iostream>using namespace std;#define MID(x,y) ((x+y)>>1)#define CLR(arr,val) memset(arr,val,sizeof(arr))#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);const double PI = acos(-1.0);const int INF = 0x3f3f3f3f;const int N=2e5+7;int main(){//freopen("f:/input.txt", "r", stdin);int zu,i,j,k,n,m,a,b,c,comb,maxx,minn;scanf("%d",&zu);while(zu--){scanf("%d%d%d",&a,&b,&c);comb=0;maxx=minn=0;for(i=0;i<c;i++){maxx+=(50*(comb*2+1));++comb;}for(i=0;i<b;i++){maxx+=(100*(comb*2+1));++comb;}for(i=0;i<a;i++){maxx+=(300*(comb*2+1));++comb;}comb = 0;for(i=0;i<a;i++){minn+=(300*(comb*2+1));++comb;}for(i=0;i<b;i++){minn+=(100*(comb*2+1));++comb;}for(i=0;i<c;i++){minn+=(50*(comb*2+1));++comb;}printf("%d %d\n",minn,maxx);}}