zoukankan      html  css  js  c++  java
  • Codeforces Gym100735 E.Restore (KTU Programming Camp (Day 1) Lithuania, Birˇstonas, August 19, 2015)

    E - Restore

    Given a matrix A of size N * N. The rows are numbered from 0 to N-1, the columns are numbered from 0 to N-1. In this matrix, the sums of each row, the sums of each column, and the sum of the two diagonals are equal.

    For example, a matrix with N = 3:

    The sums of each row:

    2 + 9 + 4 = 15

    7 + 5 + 3 = 15

    6 + 1 + 8 = 15

    The sums of each column:

    2 + 7 + 6 = 15

    9 + 5 + 1 = 15

    4 + 3 + 8 = 15

    The sums of each diagonal:

    2 + 5 + 8 = 15

    4 + 5 + 6 = 15

    As you can notice, all sums are equal to 15.

    However, all the numbers in the main diagonal (the main diagonal consists of cells (i, i)) have been removed. Your task is to recover these cells.

    Input

    The first line contains the dimension of the matrix, n (1 ≤ n ≤ 100).

    The following n lines contain n integers each, with removed numbers denoted by 0, and all other numbers  - 1012 ≤ Aij ≤ 1012.

    Output

    The restored matrix should be outputted in a similar format, with n lines consisting of n integers each.

    Example

    Input
    3
    0 9 4
    7 0 3
    6 1 0
    Output
    2 9 4 
    7 5 3
    6 1 8

     
    这个题有一个地方有点坑,就是如果一行里面有两个0,因为只有对角线的挖掉了,所以直接竖着算这一排的就可以。其他的没了。
     
    代码:
     1 #include<iostream>
     2 #include<cstring>
     3 #include<cstdio>
     4 #include<algorithm>
     5 using namespace std;
     6 typedef long long ll;
     7 ll a[105][105];
     8 int main(){
     9     int n;
    10     while(~scanf("%d",&n)){
    11             ll cnt=0;
    12         for(int i=0;i<n;i++){
    13             for(int j=0;j<n;j++){
    14                 scanf("%lld",&a[i][j]);
    15                 cnt+=a[i][j];
    16             }
    17         }
    18         cnt/=n-1;int i,j;   //算那个相等的值
    19         for(i=0;i<n;i++){
    20                 ll num=0;int flag,ret=0;
    21             for(j=0;j<n;j++){
    22                 num+=a[i][j];
    23                 if(a[i][j]==0){flag=j;ret++;}
    24             }
    25             if(ret>1){    //如果这一排有多于1个0
    26                 ll qwe=0;
    27                 for(int k=0;k<n;k++)qwe+=a[k][i];
    28                 a[i][i]=cnt-qwe;
    29             }
    30             else if(num!=cnt)a[i][flag]=cnt-num;
    31         }
    32         for(int i=0;i<n;i++){
    33             for(int j=0;j<n;j++){
    34                 printf("%lld",a[i][j]);
    35                 if(j!=n-1)printf(" ");
    36                 else printf("
    ");
    37             }
    38         }
    39     }
    40     return 0;
    41 }
  • 相关阅读:
    菜菜小问题——python中print函数 以及单引号、双引号、三引号
    fiddler软件测试——Fiddler抓取https设置详解(图文)
    【转】fiddler抓包时出现了tunnel to ......443 解密HTTPS数据
    小菜菜mysql练习解读分析2——查询存在" 01 "课程但可能不存在" 02 "课程的情况(不存在时显示为 null )
    c语言基础——基本数据类型
    linux 02
    C#中海量数据的批量插入和更新
    linux 基本命令-01
    Linux -01 安装centos
    mysql 锁和隔离事务
  • 原文地址:https://www.cnblogs.com/ZERO-/p/9703197.html
Copyright © 2011-2022 走看看