https://blog.csdn.net/csyifanZhang/article/details/105632729
为了更好的阅读体验↑

线段树比暴力更加优美，这道题能用暴力完全是出题人给面子了。。。

看评论区大佬都水过了，我也先水一下，复杂度都可以过，可以见得数据之水当然这道题的数据量本身也是符合这个复杂度的，

暴力解法

ps：fill和memset别用得太多，有时候会被卡死

#include<iostream>
#include<string>
#include<string.h>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
using namespace std;

#define ll int
#define MAX 10005
#define inf 0x3fffff

ll arr[MAX];

int main() {
    ll L, M, a, b;
    while (cin >> L >> M) {
        fill(arr, arr + MAX, 1);
        for (int i = 0; i < M; i++) {
            cin >> a >> b;
            for (int i = a; i <= b; i++)arr[i] = 0;
        }
        ll res = 0;
        for (int i = 0; i <= L; i++)res += arr[i];
        cout << res << endl;
    }
}

线段树

但是如果,？还能水过嘛，显然不能，对于区间查询，区间修改，最好的方式即线段树。这道题的线段树更新也比较简单，具体还是看

#include<iostream>
#include<string>
#include<string.h>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
using namespace std;

#define ll int
#define MAX 10005
#define inf 0x3fffff

ll t[MAX * 4], tag[MAX * 4];
inline ll lc(ll k) { return k << 1; }
inline ll rc(ll k) { return k << 1 | 1; }

void pushUpSum(ll p) { t[p] = t[lc(p)] + t[rc(p)]; }
void build(ll p, ll l, ll r){
    if (l == r) { t[p] = 1; return; }
    ll mid = (l + r) >> 1;
    build(lc(p), l, mid); 
    build(rc(p), mid + 1, r);
    pushUpSum(p);

}

inline void f(ll p, ll l, ll r, ll k) {
    tag[p] = k;
    t[p] = 0;//这里pushdown是清零操作
    if (t[p] < 0)t[p] = 0;
}
inline void pushDown(ll p, ll l, ll r) {
    ll mid = (l + r) >> 1;
    f(lc(p), l, mid, tag[p]);
    f(rc(p), mid + 1, r, tag[p]);
    tag[p] = 0;
}

inline void update(ll nl, ll nr, ll l, ll r, ll p, ll k = -1) {
    //nl-nr：需要进行-1修改的区间
    //l,r,p,k：当前节点所存储的区间以及节点的编号 
    if (nl <= l && r <= nr) {//需要修改区间包含了当前区间
        t[p] = 0; tag[p] = -1; return;//当前区间清零
    }
    if (tag[p] == -1)pushDown(p, l, r);
    ll mid = (l + r) >> 1;
    if (nl <= mid)update(nl, nr, l, mid, lc(p));
    if (nr > mid)update(nl, nr, mid + 1, r, rc(p));
    pushUpSum(p);
}

int main() {
    ll L, M, a, b;
    while (cin >> L >> M) {
        memset(t, 0, sizeof(t));
        memset(tag, 0, sizeof(tag));
        build(1, 0, L);
        for (int i = 0; i < M; i++) {
            cin >> a >> b;
            update(a, b, 0, L, 1);
        }
        cout << t[1] << endl;
    }
}