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               ####....#.
             #..###.....##....
             ###.......######              ###            ###
                ...........               #...#          #...#
               ##*#######                 #.#.#          #.#.#
            ####*******######             #.#.#          #.#.#
           ...#***.****.*###....          #...#          #...#
           ....**********##.....           ###            ###
           ....****    *****....
             ####        ####
           ######        ######
##############################################################
#...#......#.##...#......#.##...#......#.##------------------#
###########################################------------------#
#..#....#....##..#....#....##..#....#....#####################
##########################################    #----------#
#.....#......##.....#......##.....#......#    #----------#
##########################################    #----------#
#.#..#....#..##.#..#....#..##.#..#....#..#    #----------#
##########################################    ############

https://ac.nowcoder.com/acm/contest/19483/F
csdn
https://blog.csdn.net/laysan/article/details/122588791

  #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e5+10;
int f[N][10],t[10];//t中存放顺序
int n,m,l,r;


int main(){
    cin>>n>>m;
    for(int i=0;i<10;i++) f[0][i]=i; //初始状态
    for(int i=1;i<=n;i++){
        cin>>l>>r;
        for(int j=0;j<10;j++) 
            f[i][j]=f[i-1][j];//本轮初始化为上一轮
        swap(f[i][l],f[i][r]);//新一轮的改变
    }
    while(m--){
        cin>>l>>r;
        for(int i=0;i<10;i++)//l的上一轮状态
         t[f[l-1][i]]=i;//将第l-1次的序列对号至1~n

        for(int i=0;i<10;i++){//变换位置
            if(i) cout<<" "<<t[f[r][i]];
            else cout<<t[f[r][i]];
        }
        cout<<endl;
    }
    return 0;
}

https://ac.nowcoder.com/acm/contest/19483/I
https://ac.nowcoder.com/acm/contest/19483/J

#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
int a[100010];
int main()
{
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        scanf("%d", &a[i]);
    int ans = 0;
    for (int i = n; i >= 1; i--)
    {
        a[i] = a[i] - a[i - 1];
       if(a[i]>0) ans+=a[i];//前缀和差分约束思想
       //另一个样例，可以反着理解，，
       //从 n变成0
       //变成 从0变成n理解
    }
    printf("%d\n", ans);
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
//typedef __int128 LL;//2的128
typedef long long ll; //2的64
#define int long long
// 1e9,, 1e19,,1e38
int a[100010],b[100010],c[100010];
const int N=1e9+7;
void add(int n, int s[])
{
    for(int i=1;i<=n;i++)
    s[i]=(s[i]+s[i-1])%N;
}
void solve()
{
    int n,m;cin>>n>>m;
for(int i=1;i<=n;i++)
{
    a[i]=0;b[i]=0;c[i]=0;
}
while(m--)
{
    int s,pos;
    cin>>s>>pos;
    if(s==1) a[pos]++;
else if(s==2) b[pos]++;
else if(s==3)
{c[pos]++;c[pos+1]++;
}
}
add(n,a);
add(n,b),add(n,b);
add(n,c),add(n,c),add(n,c);
for(int i=1;i<=n;i++)
{
    cout<<((a[i])+(b[i])+c[i])%N<<" ";
}
    cout<<endl;
}

signed main()
{
    ios::sync_with_stdio(false);
cin.tie(0),cout.tie(0);
    int t; cin>>t;
    while(t--)
{
    solve();
}


    return 0;
}