Baidu Star Qualifying Problem 1001

Test instructions

The bear has a dictionary that stores a large number of words, and once, he makes all the words a long, long string. Now that the trouble has come, he forgets what the original string is, and what is magical is that he remembers the hash of the original string. The hash value of a string, calculated by the following formula:

Please help the bear calculates the number of hashes in any segment of a large string. Multiple sets of test data, each set of test data the first line is a positive integerN, representing the number of queries, the second line of a string, representing the large string in the title, and thenn < Span class= "Mord mathit" style= "MARGIN-RIGHT:0.10903EM;" > row, each row contains two positive integers a and B

Ideas:first, use a prefix array a[i] to represent the H (s) value of the string from the first character to the I-character, and, if required from A to B, it is obviously a[b]/a[a-1]. Also because the A array is 9973 modulo, so the equivalent is known a[b]% p, a[a-1]% p, ask A[B]/a[a-1]. known a% mod, b% mod, request (A/b)% mod, with inv (p) to represent P's inverse
(A/b)% mod = (A * INV (b))% mod = (a% mod * INV (b)% mod)% MoD;
Then ask for 1 to mod-1 mod inverse to inv (b% MoD)
So (A/b)% mod = (a% mod * INV (b% mod))% MoD;
Ask 1 to mod-1 for mod inverse code:

 const  int  mod = 9973  ; 2  LL inverse[mod + ]; 3  inverse[0 ] = 0 , Inverse[1 ] = 1   4   int  i = 2 ; i < mod; I++)  5  inverse[i] = (-(mod/i) * Inverse [MoD% i]) % mod + mod;  

This problem code:

1 Const intMAXN =1000000+7;2 Const intp =9973;3 LL ni[p];4 CharSTR[MAXN];5 LL A[MAXN];6 7 intMain () {8     //freopen ("Input.txt", "R", stdin);9ni[1] =1;Tenni[0] =0; One      for(inti =2; I < P; i++) ANi[i] = (-(p/i) * ni[p%i])% p +p; -     intN; -      while(~SCANF ("%d", &N)) the     { -scanf"%s", str); -         intLen =strlen (str); -a[0] =1; +          for(inti =1; I <= Len; i++) -A[i] = a[i-1] * (Str[i-1] - -) %p; +          while(n--){ A             ints, m; atscanf"%d%d", &s, &m); -printf"%i64d\n", A[m] * ni[a[s-1]] %p); -         } -     }  -     return 0; -}

Baidu Star Qualifying Problem 1001