给定一个具有m个整数和n的数组arr [m],n是要添加到数组中的值,并且r查询以一些开始和结尾给出。对于每个查询,我们必须从数组的限制的开始到结束为止添加值n。
Input:
arr[] = {1, 2, 3, 4, 5}
query[] = { { 0, 3 }, { 1, 2 } }
n = 2
Output:
If we run above program then it will generate following output:
Query1: { 3, 4, 5, 6, 5 }
Query2: { 3, 6, 7, 6, 5 }该程序可以通过一种简单的方法来解决,其中-
我们将迭代所有查询,从查询的起点开始遍历数组,直到存储在查询中的终点为止。
添加n的值并打印数组。
START STEP 1 : DECLARE A STRUCT range for start AND end LIMITS STEP 2 : IN FUNCTION add_tomatrix(int arr[], struct range r[], int n, int size, int m) int i, j, k; LOOP FOR i = 0 AND i < m AND i++ LOOP FOR j = r[i].start AND j<= r[i].end AND j++ arr[j] = arr[j] + n END FOR LOOP FOR k = 0 AND k < size AND k++ PRINT arr[k] END FOR END FOR STOP
#include <stdio.h>
struct range{
int start, end; //struct to give the range for the array elements
};
int add_tomatrix(int arr[], struct range r[], int n, int size, int m){
int i, j, k;
for ( i = 0; i < m; i++) //for all the elements in a struct we defined{
for(j = r[i].start; j<= r[i].end; j++) //from where till where we want our results to be updated{
arr[j] += n; //add the value of the particular range
}
printf("Query %d:", i+1);
for ( k = 0; k < size; k++){
printf(" %d",arr[k]); // print the whole array after every query
}
printf("\n");
}
}
int main(int argc, char const *argv[]){
int arr[] ={3, 4, 8, 1, 10};
struct range r[] = {{0,2}, {1, 3}, {3, 4}};
int n = 2;
int size = sizeof(arr)/sizeof(arr[0]);
int m = sizeof(r)/sizeof(r[0]);
add_tomatrix(arr, r, n, size, m);
return 0;
}输出结果
如果我们运行上面的程序,那么它将生成以下输出-
Query 1: 5 6 10 1 10 Query 2: 5 8 12 3 10 Query 3: 5 8 12 5 12