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Added solutions of Div2_Round#675
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// nb_9960
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#include <bits/stdc++.h>
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using namespace std;
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#define fi(i,a,n) for (int i=a; i < n; i++)
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#define ll long long
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#define pb push_back
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#define all(x) x.begin(), x.end()
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#define clr(x) memset(x, 0, sizeof(x))
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#define sortall(x) sort(all(x))
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typedef pair<ll, ll> pl;
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typedef vector<ll> vl;
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typedef vector<pl> vpii;
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int main()
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{
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ios_base::sync_with_stdio(0);
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cin.tie(0);
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cout.tie(0);
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ll t = 1;
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cin >> t;
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while (t--) {
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ll a,b,c;
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cin>>a>>b>>c;
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vl v={a,b,c};
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sortall(v);
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cout<<v[2]+v[1]-v[0]<<endl;
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}
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return 0;
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}
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// nb_9960
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#include <bits/stdc++.h>
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using namespace std;
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#define fi(i,a,n) for (int i=a; i < n; i++)
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#define ll long long
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#define pb push_back
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#define mp make_pair
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#define F first
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#define S second
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#define all(x) x.begin(), x.end()
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#define clr(x) memset(x, 0, sizeof(x))
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#define sortall(x) sort(all(x))
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typedef pair<ll, ll> pl;
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typedef vector<ll> vl;
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int main()
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{
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ios_base::sync_with_stdio(0);
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cin.tie(0);
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cout.tie(0);
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ll t = 1;
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cin >> t;
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while (t--) {
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ll n,m,ans=0;
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cin>>n>>m;
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vl v[n];
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fi(i,0,n){
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fi(j,0,m){
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ll x;cin>>x;
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v[i].pb(x);
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}
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}
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fi(i,0,n/2){
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fi(j,0,m/2){
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vector<ll> v1={v[i][m-j-1],v[i][j],v[n-1-i][j],v[n-1-i][m-1-j]};
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sortall(v1);
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ll mean=v1[1];
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fi(i,0,v1.size())
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ans+=abs(mean-v1[i]);
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}
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}
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if(n%2){
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fi(i,0,m/2){
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ans+=abs(v[n/2][i]-v[n/2][m-1-i]);
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}
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}
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if(m%2){
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fi(i,0,n/2){
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ans+=abs(v[i][m/2]-v[n-1-i][m/2]);
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}
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}
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cout<<ans<<endl;
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}
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return 0;
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}
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// nb_9960
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#include <bits/stdc++.h>
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using namespace std;
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#define fi(i,a,n) for (ll i=a; i < n; i++)
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#define ll long long
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#define pb push_back
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#define all(x) x.begin(), x.end()
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#define clr(x) memset(x, 0, sizeof(x))
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#define sortall(x) sort(all(x))
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typedef pair<ll, ll> pl;
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typedef vector<ll> vl;
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const ll mod=1000000007;
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ll power(ll x,ll y,ll m){if(y==0)return 1;ll p=power(x,y/2,m)%m;p=(p*p)%m;return (y%2==0)?p:(x*p)%m;}
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int main()
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{
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ios_base::sync_with_stdio(0);
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cin.tie(0);
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cout.tie(0);
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ll t = 1;
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// cin >> t;
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while (t--) {
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string s;
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cin >> s;
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ll ans=0;
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ll n=s.length();
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ll tmp=0;
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for(ll i=n-2;i>=0;--i)
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{
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ll t=(s[i]-'0');
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tmp+=(power(10,n-i-2,mod)*(n-i-1))%mod;
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tmp%=mod;
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ans+=((t*tmp)%mod);
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ans%=mod;
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}
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tmp=0;
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fi(i,1,n)
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{
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ll t=(s[i]-'0');
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tmp+=i;
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tmp%=mod;
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ans+=(((t*tmp)%mod)*power(10,n-i-1,mod))%mod;
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ans%=mod;
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}
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cout << ans << endl;
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}
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return 0;
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}

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