此文章可以使用目录功能哟↑(点击上方[+])
Kattis <Bijele>
Accept: 0 Submit: 0
Time Limit: 1 second Memory Limit : 1024 MB
Problem Description
Mirko has found an old chessboard and a set of pieces in his attic. Unfortunately, the set contains only white pieces, and apparently an incorrect number of them. A set of pieces should contain:
- One king
- One queen
- Two rooks
- Two bishops
- Two knights
- Eight pawns
Mirko would like to know how many pieces of each type he should add or remove to make a valid set.
Input
The input consists of 6 integers on a single line, each between 0 and 10 (inclusive). The numbers are, in order, the numbers of kings, queens, rooks, bishops, knights and pawns in the set Mirko found.
Output
Output should consist of 6 integers on a single line; the number of pieces of each type Mirko should add or remove. If a number is positive, Mirko needs to add that many pieces. If a number is negative, Mirko needs to remove pieces.
Sample Input
2 1 2 1 2 1
Sample Output
-1 0 0 1 0 7
Problem Idea
解题思路:
【题意】
国际象棋中有下述6种棋子:
1.一个国王
2.一个皇后
3.两个车
4.两个象
5.两个马
6.八个卒
现在给你棋盘上这6种棋子的个数,问为了使棋子数量合法,每种棋子需要增加或减少多少个
输出的数为正表示增加,为负表示减少
【类型】
暴力
【分析】
由于每种棋子的合法数量已经告诉我们,这样就避免了没有接触过国际象棋的人而无法下手的局面
所以,我们要做的就是将实际数量与合法数量一一比较就可以了
【时间复杂度&&优化】
O(6)
题目链接→Kattis <Bijele>
Source Code
/*Sherlock and Watson and Adler*/
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<bitset>
#include<cmath>
#include<complex>
#include<string>
#include<algorithm>
#include<iostream>
#define eps 1e-9
#define LL long long
#define PI acos(-1.0)
#define bitnum(a) __builtin_popcount(a)
using namespace std;
const int N = 6;
const int M = 100005;
const int inf = 1000000007;
const int mod = 10007;
int ans[N]={1,1,2,2,2,8},s[N];
int main()
{int i;for(i=0;i<N;i++){scanf("%d",&s[i]);printf("%d%c",ans[i]-s[i],i!=N-1?' ':'\n');}return 0;
}