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  1. #include <bits/stdc++.h>
  2. #include <ext/pb_ds/assoc_container.hpp>
  3. #include <ext/pb_ds/tree_policy.hpp>
  4.  
  5. #define ll long long
  6. #define fi(i,l,n)   for(int i=l; i<n; i++)
  7. #define fd(i,n,l)   for(int i=n; i>=l; i--)
  8. #define gets(s) string s; cin>>s;
  9. #define getv(v,n)   vector<ll> v(n+1); fi(i,1,n+1)    cin>>v[i];
  10. #define all1(v)  v.begin() + 1, v.end()
  11. #define all(v)  v.begin(), v.end()
  12. #define seev(v,n) fr(i,1,n+1)   {cout<<v[i]<<' ';}  cout<<endl;
  13. #define vl vector<ll>
  14. #define vvl vector<vector<ll>>
  15. #define vp vector<pair<ll,ll>>
  16. #define vc vector<char>
  17. #define vvc vector<vector<char>>
  18.  
  19. #define pb push_back
  20. #define mp make_pair
  21. #define mse multiset<ll>
  22. #define se set<ll>
  23. #define ma map<ll,ll>
  24. #define getmat(v,n,m) vvl v(n+1,vl(m+1)); fi(i,1,n+1) {fi (j,1,m+1) cin>>v[i][j];}
  25. #define seemat(mat) fi(i,1,n+1) {fi (j,1,m+1) cout<<v[i][j]<<' '; cout<<endl;}
  26. #define YES cout<<"YES"<<endl;
  27. #define NO cout<<"NO"<<endl;
  28. using namespace std;
  29. using namespace __gnu_pbds;
  30.  
  31. // Define ordered multiset with long long
  32.  
  33. typedef tree<long long, long long, less<long>, rb_tree_tag, tree_order_statistics_node_update> ordered_multimap;
  34. typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
  35.  
  36. vl sieve;
  37. void SieveOfEratosthenes(int n)
  38. {
  39.     // Create a boolean array "prime[0..n]" and initialize
  40.     // all entries it as true. A value in prime[i] will
  41.     // finally be false if i is Not a prime, else true.
  42.     vl prime(n+1,true);
  43.  
  44.     for (int p = 2; p * p <= n; p++) {
  45.         // If prime[p] is not changed, then it is a prime
  46.         if (prime[p] == true) {
  47.             // Update all multiples of p greater than or
  48.             // equal to the square of it numbers which are
  49.             // multiple of p and are less than p^2 are
  50.             // already been marked.
  51.             for (int i = p * p; i <= n; i += p)
  52.                 prime[i] = false;
  53.         }
  54.     }
  55.  
  56.     // Print all prime numbers
  57.     for (int p = 2; p <= n; p++)
  58.         if (prime[p])
  59.             sieve.push_back(p);
  60. }
  61.  
  62. class DisjointSet{
  63.     private:
  64.     vector<int> parent;
  65.  
  66.     public:
  67.     DisjointSet(int n){
  68.         parent.resize(n+1,-1);
  69.     }
  70.  
  71.     int findParent(int x){
  72.         if(parent[x] < 0) return x;
  73.         return parent[x] = findParent(parent[x]);
  74.     }
  75.  
  76.     void Union(int x, int z){
  77.         int px = findParent(x);
  78.         int pz = findParent(z);
  79.  
  80.         if(px == pz)    return;
  81.  
  82.         if(parent[px] < parent[pz]){
  83.             parent[px] += parent[pz];
  84.             parent[pz] = px;
  85.         }
  86.         else {
  87.             parent[pz] += parent[px];
  88.             parent[px] = pz;
  89.         }
  90.  
  91.     }
  92.  
  93.     int getSize(int x){
  94.         int px = findParent(x);
  95.         return ((-1)*parent[px]);
  96.     }
  97. };
  98.  
  99. bool sortbysec(const pair<ll, ll>& a, const pair<ll, ll>& b){
  100.     return (a.second < b.second);
  101. }
  102.  
  103.  
  104. ll gcd(ll a, ll b){
  105.     if (b == 0) return a;
  106.     return gcd(b,a%b);
  107. }
  108.  
  109.  
  110. ll pwr(ll a, ll b, ll mod = 0){
  111.     ll result = 1;
  112.     if(mod == 0){
  113.         while(b){
  114.             if(b & 1) result *= a;
  115.             a *= a;
  116.             b = b >> 1;
  117.         }
  118.     }
  119.     else {
  120.         while(b){
  121.             if(b & 1){
  122.                 result *= a;
  123.                 result  = result % mod;
  124.             }
  125.             a *= a;
  126.             a = a % mod;
  127.             b = b >> 1;
  128.         }
  129.     }
  130.     return result;
  131. }
  132.  
  133.  
  134. ll modularinverse(ll a,ll p)// p is prime
  135. {
  136.     return pwr(a,p-2,p);
  137. }
  138.  
  139. void solve(ll tc){
  140.     int n,m;
  141.     cin>>n>>m;
  142.  
  143.     string s;
  144.     cin>>s;
  145.  
  146.     ll a[n][m];
  147.  
  148.     fi(i,0,n) fi(j,0,m) cin>>a[i][j];
  149.  
  150.     int r=0,c=0;
  151.     int step = 0;
  152.     a[r][c] = 0;
  153.  
  154.     ll sum=0;
  155.  
  156.     if(s[0] == 'D'){
  157.         fi(col,0,m) sum+=a[0][col];
  158.     }
  159.     else{
  160.         fi(row,0,n) sum+=a[row][0];
  161.     }
  162.  
  163.     a[r][c] = -1*sum;
  164.     sum=0;
  165.  
  166.  
  167.  
  168.     while(step<n+m-1){
  169.         ll curr=0;
  170.         step++;
  171.  
  172.         if(s[step-1] == 'D'){
  173.             r++;
  174.         }
  175.         else c++;
  176.  
  177.         if(s[step] == 'D'){
  178.             fi(col,0,m){
  179.                 if(col == c) continue;
  180.                 curr+=a[r][col];
  181.             } 
  182.  
  183.             a[r][c] = sum - curr;
  184.         }
  185.         else{
  186.             fi(row,0,n){
  187.                 if(row == r) continue;
  188.                 curr+=a[row][c];
  189.             } 
  190.  
  191.             a[r][c] = sum - curr;
  192.         }
  193.  
  194.     } 
  195.  
  196.  
  197.     // ll curr = 0;
  198.  
  199.     fi(i,0,n){
  200.         fi(j,0,m) cout<<a[i][j]<<' ';
  201.         cout<<endl;
  202.     }
  203.  
  204.     // fi(i,0,n){
  205.     //     fi(j,0,m) curr+=a[i][j];
  206.     //     if(curr!= 0 ) NO
  207.     //     curr = 0;
  208.     // }
  209.  
  210.     // fi(j,0,m){
  211.     //     fi(i,0,n) curr += a[i][j];
  212.     //     if(curr != 0) NO
  213.     //     curr = 0;
  214.     // }
  215.     // YES
  216. }
  217.  
  218. int main(){
  219.     ios_base::sync_with_stdio(false);
  220.     cin.tie(NULL);
  221.     cout.tie(NULL);
  222.  
  223.     int t=1;
  224.     cin>>t;
  225.     ll tc=0;
  226.     while(t--){
  227.         tc++;
  228.         solve(tc);
  229.     }
  230.  
  231.     return 0;
  232. }
Success #stdin #stdout 0s 5288KB
comments (?)
stdin 
4
3 3
DRRD
0 2 3
0 0 0
3 1 0
4 5
DRRRRDD
0 1 0 2 3
0 0 0 0 0
-1 0 -3 -3 0
0 0 0 -1 0
2 3
RRD
0 0 0
0 1 0
5 5
DDDDRRRR
0 25 2 9 11
0 6 13 20 22
0 17 24 1 8
0 3 10 12 19
0 0 0 0 0
stdout 
-5 2 3 
2 -3 1 
3 1 -4 
-6 1 0 2 3 
7 -1 3 2 -11 
-1 0 -3 -3 7 
0 0 0 -1 1 
0 -1 1 
0 1 -1 
-47 25 2 9 11 
-61 6 13 20 22 
-50 17 24 1 8 
-44 3 10 12 19 
202 -51 -49 -42 -60
https://ideone.com/gos1we
language:
C++ (gcc 8.3)
created:
2 days ago
visibility:
public

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