The road off the east gate of Peking University used to be decorated with a lot of trees. However, because of the construction of a subway, a lot of them are cut down or moved away. Now please help to count how many trees are left.  Let's only consider one side of the road. Assume that trees were planted every 1m (meter) from the beginning of the road. Now some sections of the road are assigned for subway station, crossover or other buildings, so trees in those sections will be moved away or cut down. Your job is to give the number of trees left.  For example, the road is 300m long and trees are planted every 1m from the beginning of the road (0m). That's to say that there used to be 301 trees on the road. Now the section from 100m to 200m is assigned for subway station, so 101 trees need to be moved away and only 200 trees are left.

There are several test cases in the input. Each case starts with an integer L (1 <= L < 2000000000) representing the length of the road and M (1 <= M <= 5000) representing the number of sections that are assigned for other use.  The following M lines each describes a section. A line is in such format:  Start End  Here Start and End (0 <= Start <= End <= L) are both non-negative integers representing the start point and the end point of the section. It is confirmed that these sections do not overlap with each other.  A case with L = 0 and M = 0 ends the input.

Output the number of trees left in one line for each test case.

300 1100 200500 2100 200201 3000 0

200300

一个L（L<2 * 10^9）长度的公路，上面每隔一米有一棵树，即这个公路上有L+1棵树。之后有M个区间的树要移除，区间为[start,end]，所以每次会移除end-start+1棵树。问最后还剩下多少棵树。

用一个unsigned int储存总数，然后每次对于区间减去需要移除的数量，最后求值。

注意点：

1）unsigned int为32位，最大值为4294967296，可以存储的下。

代码（1AC， POJ100题啦 \(^o^)/）：

#include 
    
     #include 
     
      #include 
      
       int main(void){    unsigned int trees;    int i, j;    int M;    unsigned int start, end;    while (scanf("%u%d", &trees, &M), trees != 0 || M != 0){        trees += 1;        for (i = 0; i < M ; i++){            scanf("%u%u", &start, &end);            trees -= end - start + 1;        }        printf("%u\n", trees);    }    return 0;}