给定一个数组 arr[m]，其中包含 m 个整数和 n（要添加到数组中的值），并给出 r 个查询，并给出一些开始和结束。对于每个查询，我们必须在数组中添加从开头到限制末尾的值 n。
示例
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 }这个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。
- 添加 n 的值并打印数组。
算法
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示例
#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("</p><p>");
}
}
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
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
