var
  l,i,j,m,sum:integer;
  st,en:array[0..100] of integer;
  c:array[0..10000] of boolean;
begin
  sum:=0;
  read(l,m);
  for i:=1 to m do
    begin
      read(st[i]);
      readln(en[i]);
    end;
  for i:=0 to l do
    c[i]:=true;
  for i:=1 to m do
    for j:=st[i] to en[i] do
      c[j]:=false;
  for i:=0 to l do
    if c[i] then
      inc(sum);
  writeln(sum);
end.

数组模拟就好了
#include <cstdio>
#include <iostream>
using namespace std;
int L, N, i, x, y;
int A[10001] = {0};
int ans=0, j;
int main() {
cin>>L>>N;
for(i=0; i<N; i++) {
cin>>x>>y;
for(j=x; j<=y; j++) {
A[j] = 1;
}
}
for(i=0; i<=L; i++)
if(A[i] == 0) ans++;
cout<<ans;
return 0;
}

#include <iostream>

using namespace std;

bool road[10001];
int mapb[101],mape[101],n,p,x;

int main()
{
int i,j;
cin>>n>>p;
x=n+1;
for(i=0;i<p;++i)
cin>>mapb[i]>>mape[i];
for(i=0;i<p;++i)
for(j=mapb[i];j<=mape[i];++j)
if(!road[j])road[j]=true,x--;
cout<<x;
return 0;
}

lishi

本题为继陶陶摘苹果后noip历时第二水题
我61秒就写完啦哈哈

#include<iostream>
using namespace std;
int a[101],b[101],l[10001];
int main()
{
    int lon,m,sum=0;
    cin>>lon>>m;
    for(int i=0;i<=lon;i++) l[i]=1;
    for(int i=0;i<m;i++) cin>>a[i]>>b[i];
    for(int i=0;i<m;i++)
    {
        for(int k=a[i];k<=b[i];k++)    l[k]=0;
    }
    for(int i=0;i<=lon;i++)
    {
        if(l[i]==1)
        sum++;
    }
    cout<<sum;
    return 0;
}

水题，差分可秒杀
```c++
#include <bits/stdc++.h>
using namespace std;

int L[10005], l, m, a, b;

int main()
{
memset(L, 0, sizeof L);
scanf("%d%d", &l, &m);
for (int i = 1; i <= m; i++) {
scanf("%d%d", &a, &b);
L[a]++;
L[b+1]--;
}
int ans = 0;
for (int i = 0; i <= l; i++) {
if (i) L[i] += L[i-1];
if (L[i] == 0)
ans++;
}
cout << ans << endl;
return 0;
}

线段树
```c++
#include <bits/stdc++.h>
#define maxN 10010

using namespace std ;
typedef long long QAQ ;

struct Tree {int l , r ;QAQ sum ,dec;};

Tree tr[maxN<<2];

void Build_Tree ( int x ,int y , int i ){
tr[i].l = x ;
tr[i].r = y ;
if(x==y)tr[i].sum = 1 ;
else {
QAQ mid = ( tr[i].l + tr[i].r ) >> 1 ;
Build_Tree ( x , mid , i<<1);
Build_Tree ( mid+1 , y , i<<1|1);
tr[i].sum = tr[i<<1].sum + tr[i<<1|1].sum ;
}
}

void Update_Tree ( int q , int w , int i){
if(!tr[i].sum)return ;
if( q<=tr[i].l && w>=tr[i].r )tr[i].sum = 0 ;
else {
QAQ mid = (tr[i].l + tr[i].r ) >> 1 ;
if ( q>mid )Update_Tree ( q , w , i<<1|1);
else if ( w<=mid )Update_Tree ( q , w , i<<1);
else{
Update_Tree ( q , w , i<<1|1);
Update_Tree ( q , w , i<<1);
}
tr[i].sum = tr[i<<1|1].sum + tr[i<<1].sum ;
}
}

int main ( ){
int N,M,l,r;
scanf("%d %d",&N,&M);
Build_Tree ( 0 , N , 1 );
while ( M-- ){
scanf("%d %d",&l,&r);
Update_Tree ( l , r , 1 );
}
printf("%I64d",tr[1].sum);
return 0 ;
}
```

数据太弱了不过我们还是写一个强一点的代码吧。codevs上可过500000的n和m数据
```c++
测试数据 #0: Accepted, time = 0 ms, mem = 2512 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 2512 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 2508 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 2512 KiB, score = 10
测试数据 #4: Accepted, time = 15 ms, mem = 2512 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 2512 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 2508 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 2508 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 2508 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 2516 KiB, score = 10
Accepted, time = 15 ms, mem = 2516 KiB, score = 100
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std;

#define maxn 500000+10
int fa[maxn];
int n, q;
int ans ;

int solve(int a,int b)
{
int i=b;
for(int i = b ; i >= a ; )//从右向左扫描
{
if( fa[i] == i )
{
ans--;//切了一个
fa[i] = a - 1;//标记此水果已切，同时用fa[i]来方便下次快速转换位置
//即直接免去查找此区间，跳到区间的一个左端点
i--;
}
else//已经扫过
i = fa[i];//直接跳到第一次切这个水果的左区间端点向右一个位置（因为已经扫过）
}
}

int main( )
{
scanf("%d %d",&n,&q);
for(int i = 0 ; i <= n ; ++i)
fa[i] = i;
ans = n+1;
for(int i = 1 ; i <= q ; i++)
{
int a, b;
scanf("%d %d",&a,&b);
if( a > b )
swap( b, a );
solve( a, b );
}
printf("%d\n",ans);
}
```

编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 596 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 600 KiB, score = 10

测试数据 #2: Accepted, time = 15 ms, mem = 600 KiB, score = 10

测试数据 #3: Accepted, time = 0 ms, mem = 600 KiB, score = 10

测试数据 #4: Accepted, time = 0 ms, mem = 600 KiB, score = 10

测试数据 #5: Accepted, time = 15 ms, mem = 600 KiB, score = 10

测试数据 #6: Accepted, time = 0 ms, mem = 600 KiB, score = 10

测试数据 #7: Accepted, time = 0 ms, mem = 596 KiB, score = 10

测试数据 #8: Accepted, time = 0 ms, mem = 596 KiB, score = 10

测试数据 #9: Accepted, time = 0 ms, mem = 600 KiB, score = 10

Accepted, time = 30 ms, mem = 600 KiB, score = 100

代码

#include<iostream>
using namespace std;
int a[10000]={0};
int main()
{
int i,j,k=0,l,m,x,y;
cin>>l>>m;
for(i=0;i<m;i++)
{
cin>>x;
cin>>y;
for(j=x;j<=y;j++)
a[j]=1;
}
for(i=0;i<=l;i++)
if(a[i]==0)
k++;
cout<<k;
return 0;
}

int i,j,k,l,n,m;
cin>>j>>k;
for(i=1;i<=k;i++){
cin>>l>>n;
m=n-l+1;
j-=m;
}
cout<<j;

program vijosP1103;

type   segment_t=record
                l,r:longint;
                sl,sr:longint;
                data:boolean;
                mid:longint;
        end;

var
        t:array[1..20005] of segment_t;
        size:longint;
        long,m:longint;
        tmp:longint;

procedure init;
begin
        t[1].l:=0;
        t[1].r:=long;
        size:=1;
end;

procedure build(id:longint);
begin
        t[id].data:=false;
        if t[id].l=t[id].r then
                exit;
        t[id].mid:=(t[id].l+t[id].r) div 2;

        inc(size);
        t[id].sl:=size;
        t[size].l:=t[id].l;
        t[size].r:=t[id].mid;
        build(size);

        inc(size);
        t[id].sr:=size;
        t[size].l:=t[id].mid+1;
        t[size].r:=t[id].r;
        build(size);
end;

procedure insert(s,d,id:longint);
begin
        if (s=t[id].l) and (d=t[id].r) then
        begin
                t[id].data:=true;
                exit;
        end;
        if (s>t[id].mid) then
        begin
                insert(s,d,t[id].sr);
                exit;
        end;

        if (d<t[id].mid+1) then
        begin
                insert(s,d,t[id].sl);
                exit;
        end;

        insert(s,t[id].mid,t[id].sl);
        insert(t[id].mid+1,d,t[id].sr);
end;

procedure query(id:longint);
begin
        if t[id].data=true then
        begin
                inc(tmp,t[id].r-t[id].l+1);
                exit;
        end;
        if t[id].l=t[id].r then exit;
        query(t[id].sl);
        query(t[id].sr);
end;

procedure inputdata;
begin
        readln(long,m);
end;

procedure work1;
var
        i:longint;
        s,d:longint;
begin
        for i:=1 to m do
        begin
                readln(s,d);
                insert(s,d,1);
        end;
end;

procedure work2;
begin
        query(1);
        writeln(long-tmp+1);
end;

begin
        tmp:=0;
        inputdata;
        init;
        build(1);
       work1;
        work2;
end.

#include<stdio.h>

int main()
{
    int l,m,i,j;
    int arr[10001]={0};
    int area[100][2];
    scanf("%d %d",&l,&m);
    for(i=0;i<m;i++)
    {
        scanf("%d %d",&area[i][0],&area[i][1]);
    }
    for(i=0;i<m;i++)
        for(j=area[i][0];j<=area[i][1];j++)
            arr[j]=1;
    j=0;
    for(i=0;i<=l;i++)
        if(arr[i]==0)
            j++;
    printf("%d",j);
    return 0;
}

var
a:array[0..20000]of boolean;
b:array[1..200]of longint;
c:array[1..200]of longint;
l,m,i,j,n:longint;
begin
readln(l,m);
fillchar(a,sizeof(a),true);
n:=0;
for i:=1 to m do readln(b[i],c[i]);
for i:=1 to m do
for j:=b[i] to c[i] do
a[j]:=false;
for i:=0 to l do if a[i]=true then inc(n);
writeln(n);
end.

#include<stdio.h>
int tr[10005];
void del(int a, int b) {
for(int i = a; i <= b; i++)
tr[i] = 0;
return;
}
int main() {
int l, m, tot = 0;
int a[10], b[10];
while(scanf("%d%d", &l, &m)) {
for(int i = 1; i <= l; i++)
tr[i] = 1;
for(int i = 0; i < m; i++) {
scanf("%d%d", &a, &b);
del(a[i], b[i]);
}
for(int i = 1; i <= l; i ++) {
if(tr[i] == 1) tot++;
}
printf("%d", tot);
}
}

每次将移走区域的数全部设为false，最后剩下的累加，然后输出
var
a:array [0..10000] of boolean;
i,m,n,d,j,o,k:integer;
begin
readln(d,o);
fillchar(a,sizeof(a),true);
for i:=1 to o do

begin

readln(m,n);
for j:=m to n do a[j]:=false;
end;
for i:=0 to d do
if a[i] then k:=k+1;
writeln(k);
end.

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <algorithm>
#include <math.h>

using namespace std;

int l,m;
bool a[10005];

int main(){
int c;
memset(a,1,sizeof(a));
int tmp1,tmp2;
scanf("%d%d",&l,&m);
int cnt = l + 1;
for(int i = 1;i <= m;i ++){
scanf("%d%d",&tmp1,&tmp2);
for(int j = tmp1;j <= tmp2;j ++){
if(a[j]){
a[j] = 0; cnt --;
}
}
}
printf("%d",cnt);
return 0;
}

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <algorithm>
#include <math.h>

using namespace std;

int l,m;
bool a[10005];

int main(){
int c;
memset(a,1,sizeof(a));
int tmp1,tmp2;
scanf("%d%d",&l,&m);
for(int i = 1;i <= m;i ++){
scanf("%d%d",&tmp1,&tmp2);
for(int j = tmp1;j <= tmp2;j ++){
a[j] = 0;
}
}
int cnt = 0;
for(int i = 0;i <= l;i ++){
if(a[i]) cnt ++;
}
printf("%d",cnt);
return 0;
}

#include<iostream>

typedef bool* Arraybol;
typedef int Temp_int;

int main(int argc,char** argv)
{
using std::cin;
using std::cout;
int n,m,i,j,ans;
cin>>n>>m;
ans=n+1;
Arraybol a=new bool[n];
bool** p=new bool;
for(p=a;p!=a+n;p++)
*p=true;
for(int i=1;i<=m;i++)
{
Temp_int x,y;
cin>>x>>y;
for(int j=x;j<=y;j++)
if(a[j])
{
ans--;
a[j]=false;
}
}
cout<<ans<<'\n';
return 0;
}

main()
{
int a[10001]={0};
int l,m;
int x;
int i,j;
int n1,n2;
scanf("%d%d",&l,&m);

for(i=1;i<=m;i++)
{
scanf("%d%d",&n1,&n2);
for (j=n1;j<=n2;j++)
a[j]=1;
}
x=0;
for(i=0;i<=l;i++)
if(a[i]==0)
x++;
printf("%d",x);
}
数组定义10000会出错好几个点过不了= =

var
p:array[0..10000] of integer;
x,y:array[1..100] of integer;
j,e,l,m,g:integer;
begin
readln(l,m);
for j:=1 to m do
readln(x[j],y[j]);
for j:=0 to l do
p[j]:=1;
for e:=1 to m do
for j:=x[e] to y[e] do
p[j]:=0;
g:=0;
for e:=0 to l do
if p[e]=1 then inc(g);
writeln(g);
end.