Nine degrees OJ 1088 remaining trees

Topic 1088: The remaining trees

time limit:1 seconds

Memory limit:32 MB

Special question: No

submitted:4253

Resolution:1907

Title Description:

There is a length of the whole number of L (1<=l<=10000) of the road, you can imagine the length of the axis of a line of L, the starting point is the origin of the coordinates, at each integer coordinate point there is a tree, that is, in 0,1,2,...,l a total of l+1 positions have l+1 tree.
Now to remove some trees, the interval of the removed tree is represented by a pair of numbers, such as 100 200 to remove all trees from 100 to 200 (including the endpoints).
There may be an M (1<=m<=100) interval, and there may be overlapping between intervals. The number of trees remaining after the tree is now required to remove all intervals.

Input:

Two integer L (1<=l<=10000) and M (1<=m<=100).
Next there are M-group integers, each with a pair of numbers.

Output:

There may be multiple sets of input data, and for each set of input data, a number is output that represents the number of trees left after the tree has been removed from all intervals.

Sample input:

500 3100 200150) 300470 471

Sample output:

298

#include <stdio.h> #include <string.h>int a[10001];int main (int argc, char *argv[]) {    int l,m;    int start,end;    while (~SCANF ("%d%d", &l,&m))    {        memset (a,0,sizeof (a));        for (int i=0;i<m;++i)        {            scanf ("%d%d", &start,&end);            for (int i=start;i<=end;++i)                a[i]=1;        }        int ans=0;        for (int i=0;i<=l;++i)            if (a[i]==0)                ans++;        printf ("%d\n", ans);    }    return 0;}

Nine degrees OJ 1088 remaining trees