1031: Find the law I

Time Limit: 1 Sec Memory Limit: 128 MB
Commits: 417 Resolved: 180
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Topic description

The old Chinese medicine doctor Bei Lei gave the big bread an n*m matrix, and the mysterious clothes that solve this matrix (add up and output each element of this matrix) will never fall to the bottom of the bed again (the big bread clothes something always falls to the bottom of the bed).

The old Chinese doctor gave some samples of the big bread, please help the big bread to find out the law and crack the mystery.

5 6

2 3 4 5 6 7

3 2 5 3 7 4

4 5 2 7 8 3

5 3 7 2 9 5

6 7 8 9 2 11

154

3 10

2 3 4 5 6 7 8 9 10 11

3 2 5 3 7 4 9 5 11 6

4 5 2 7 8 3 10 11 4 13

187

8 8

2 3 4 5 6 7 8 9

3 2 5 3 7 4 9 5

4 5 2 7 8 3 10 11

5 3 7 2 9 5 11 3

6 7 8 9 2 11 12 13

7 4 3 5 11 2 13 7

8 9 10 11 12 13 2 15

9 5 11 3 13 7 15 2

442

enter

Multiple sets of data

Each set of data contains one row, 2 numbers N, M, N represents the number of rows, M represents the number of columns (1 <= N, M <= 1000)

output

Output a row, the sum of all elements of the matrix

sample input

5 6
3 10
8 8

Sample output

154
187
442 

Analysis: The problem of finding the greatest common divisor 
　　At first glance, this question has no clue. It seems that there is no regularity between the rows and columns of the matrix. After searching on the Internet, I found that this regularity is really difficult to find. 
　　Looking at the first row of each matrix, it is not difficult to find that it increases from 2 to 2+M-1. Take the second input as an example to observe the remaining rows: 
　　　
　　　j: 1 2 3 4 5 6 7 8 9 10 
　　i= 2: 3 2 5 3 7 4 9 5 11 6 
　　i+j: 4 4 5 6 7 8 9 10 11 12 
　　i=3: 4 5 2 7 8 3 10 11 4 13 
　　i+j: 4 5 6 7 8 9 10 11 12 13 
　　
　　and so on, it is not difficult to find that when i and j do not have the greatest common divisor, the corresponding value of the input matrix is (i+j), and when there is the greatest common divisor, the corresponding value of the input matrix is (i+j)/ i, the greatest common divisor of j. 
　　The code is as follows: 
　　　　#include<iostream> 
　　　　using namespace std; 
　　　　//The efficiency of tossing and dividing is high 
　　　　int gcd(int a,int b){ 
 　　　　　　if(b==0) 
 　　　　　　　　 return a; 
 　　　　　　return gcd(b,a%b); 
　　　　} 
　　　　int main(){ 
 　　　　　　int N,M,ans;
　　　　　　 while(scanf("%d%d",&N,&M)!=EOF){ 
 　　　　　　　　 ans=0; 
  　　　　　　　　for(int i=1;i<=N;i++){ 
   　　　　　　　　　　for(int j=1;j<=M ;j++){ 
    　　　　　　　　　　　　ans+=(i+j)/gcd(i,j); 
   　　　　　　　　　　} 
  　　　　　　　　　} 
  　　　　　　　　printf("%d\n",ans); 
 　　　　　　　} 
　　　　} 
　　By the way, enter multiple sets of data while(scanf("%d %d",&N,&M)!=EOF), if EOF is not added, the time limit will be exceeded.