Professor Zhou's Big Data Template [full version]

/* This was written by Professor Zhou shangchao many years ago. I hope to read this article and remember that such a mathematician has been quietly contributing !! Let's give a brief introduction to our professors. during the Cultural Revolution, we learned foreign languages, mastered English, met Japanese, Russian, German, and French Foreign Languages, and early excellent intellectuals, A young man who has not yet graduated from Wuhan University on his own in 1983 and is engaged in mathematics after graduation. He has published various academic articles in a number of foreign academic newspapers and has made great achievements in his studies, with the honor of an International mathematician, we have brought out a large number of outstanding talents .... I was lucky to have dinner at the professor's house last time. Thank you very much... today is Christmas. I hope the professor's health will be better... it is also a kind of happiness for students who can become professors in their lives... * // * high-precision computing template function 2011-9-5 file name bigint. the H integer is greater than 2 ^ 64-1. It is called a big integer and must be computed with high precision. For example, if a = 123444455556666777788889999 is a 27-digit integer, it is represented by an array. The minimum 4 digits are represented by a [1] = 9999, A [1] is the single digit of the 10000 hexadecimal number. A [2] = 8888; A [3] = 7777, a [4] = 6666, a [5] = 5555, a [6] = 4444, A [7] = 123; A [0] = 7; indicates that it is the 7-digit number of the hexadecimal number. Base = 10000 hexadecimal. Large integers are read in strings, and then converted into Arrays Using the tran function. The following functions are available in this file: Trans, comp, Pn, copy, add, sub, mult, mult, Div, Div, when high-precision computing is used in the GCD Program, (1) copy this file to the front of int main (), and delete unnecessary functions. (2) You can also add # include <bigint. h> for example, hdu1002 A + B Problem II. You can only use addition to delete sub, mult, mult, and Div. You can use hdu1002 A + B problem iihdu1502 regular wordshdu1042 N for high-precision calculation! N <= inherit hdu1133 buy the ticket http://acm.hdu.edu.cn/showproblem.php? PID = 1133zju2061 http://acm.zju.edu.cn/onlinejudge/showProblem.do? Problemcode = 2061hdu1134 game of connections catalan number H [1] = 1, h [N] = H [n-1] * (4n-2)/(n + 1 ), n <= 100 hdu1250 hat's Fibonacci http://acm.hdu.edu.cn/showproblem.php? PID = 1250pku2084 game of connections = hdu1134pku2515 birthday cake calculation 1 + 2 ^ m +... + N ^ m, n <= 10 ^ 41, m <= 100pku1220 number base conversionpku2680 computer hour dollar dayzpku3199 uncle jackpku3324 Lucas-Lehmer testpku3331 the idiot Of The Year contest */# include <vector> # include <list> # include <map> # include <set> # include <deque> # include <stack> # include <bitset> # include <algorithm> # include <Functional> # include <numeric> # include <utility> # include <sstream> # include <iostream> # include <iomanip> # include <cstdio> # include <cmath> # include <cstdlib> # include <ctime> # include <stdio. h> # include <string. h> using namespace STD; const int base = 10000; // The maximum value of a hichina Int. If you only use +,-, you can increase the value of base = 000000000int one [] = {1, 1 }; // large integer 1int zero [] = {1, 0}; // large integer 0int comp (int * a, int * B ); // compare the large integer int comp (int * a, int * B, int T); // Large integer compare void trans (char * Ch, int * A); // string-> Number base = 10000, a [1] is a percentile, A [0] is the number of digits (segment number) void Pn (int * A); // printf large integer avoid copy (int * a, int * B ); // A = bvoid add (int * a, int * B, int * C); // large integer + large integer c = a + bvoid sub (int *, int * B, int * C); // big integer a-big integer B, A> bvoid mult (int * a, int B, int * D ); // big integer * integer c = A * bvoid mult (int * a, int * B, int * C ); // big integer * big integer c = A * bint Div (int * a, int B, int * C ); // C = A/B k = A % bvoid Div (int * a, int * B, int * C, Int * r); // C = A/B r = A % bvoid gcd (int * a, int * B, int * D); // d = gcd (, b)/* string --> Number base = 10000, a [1] is the second bit, And a [0] is the number of digits (segment number) the string is 12355558888 --> A [1] = 8888, A [2] = 5555, a [3] = 123 A [0] = 3, each segment seg = 4 digits */void trans (char * Ch, int *) {int L, I, J, K, S, I9, SEG = 4; // seg = 6; // base = 1000000 L = strlen (CH ); S = L-seg; k = 1; for (I = s; I> = 0; I-= seg, K ++) {I9 = I + seg; A [k] = CH [I]-'0'; For (j = I + 1; j <I9; j ++) A [k] = A [k] * 10 + CH [J]-48;} I + = seg; A [k] = 0; For (j = 0; j <I; j ++) A [k] = A [k] * 10 + CH [J]-48; if (a [k]) A [0] = K; else a [0] = K-1;} // large integer compare the size of int comp (int * a, int * B) {int I; if (A [0]> B [0]) return 1; if (a [0] <B [0]) Return-1; for (I = A [0]; i> = 1; I --) {if (a [I]> B [I]) return 1; if (a [I] <B [I]) return-1;} return 0;} int comp (int * a, int * B, int t) {int I, j = 1; for (I = B [0]; i> = 1; I --, J ++) {if (a [T]> B [I]) return J; if (a [T] <B [I]) return-1; t --;} return 0;} void PN (Int * A) {int I; // output "% 04d" 10000 hexadecimal printf ("% d", a [A [0]); for (I = A [0]-1; I> = 1; I --) printf ("% 04d", a [I]); printf ("\ n");} void copy (int * a, int * B) {int I; // A = B for (I = 0; I <= B [0]; ++ I) A [I] = B [I];} // large integer + large integer c = a + B, base hexadecimal, base = 10000 // C [0] indicates the number of segments, and C [1] indicates the percentile bit. // Example 9912348888, C [0] = 3, C [1] = 8888, C [2] = 1234, C [3] = 99 void add (int *, int * B, int * c) {int S, I, T, P, d [1000]; If (B [0] = 1) & (B [1] = 0) {copy (C, A); return;} If (A [0] = 1) & (A [1] = 0) {copy (C, B); return;} if (a [0]> = B [0]) {copy (C, A); copy (D, B) ;}else {copy (C, B); copy (D, a) ;}s = C [0]; T = d [0]; // C [0]> = d [0] C [S + 1] = 0; for (I = 1; I <= T; I ++) {C [I] + = d [I]; If (C [I]> = base) {C [I]-= base; c [I + 1] ++ ;}}for (; I <= s; I ++) {If (C [I]> = Base) {c [I]-= base; C [I + 1] ++;} else break;} If (C [S + 1]> 0) c [0] = S + 1; // process the last digit} // large integer a-big integer B, A> B, C = A-bvoid sub (int * a, int * B, int * c) {int I, j, P = 0; if (B [0] = 1 & (B [1] = 0) {copy (C, A); return;} If (COMP (A, B) <= 0) {copy (C, zero); return;} If (COMP (A, zero) = 0) {copy (C, zero); return ;} for (I = 1; I <= B [0]; I ++) {C [I] = A [I]-B [I]-P; if (C [I] <0) {C [I] + = base; P = 1;} else p = 0;} For (; I <= A [0]; I ++) {C [I] = A [I]-P; If (C [I] <0) {C [I] + = base; P = 1 ;} else p = 0 ;}for (I = A [0]; I> = 1; I --) {If (C [I]) {C [0] = I; break ;}}// big integer a multiply the constant B <base, D = A * bvoid mult (int * a, int B, int * D) {int W, I, p; If (B = 1) {copy (D, a); return;} If (B = 0) | (A [0] = 1 & A [1] = 0) {d [0] = 1; d [1] = 0; return ;} W = a [0]; P = 0; // for (I = 1; I <= W; I ++) {d [I] = A [I] * B + P; If (d [I]> = base) {P = d [I]/base; d [I] % = base;} else p = 0;} If (p) {W ++; d [w] = P;} d [0] = W ;} // big integer a * big integer B, C = A * bvoid mult (int * a, int * B, int * c) {int I, j, S, m, n, K, P; // base <= 10000 m = A [0]; n = B [0]; k = m + n-1; /* m digits * n digits = m + n-1 digits or m + n digits instance 12345678*123456789012 = 1524157764056090136 a [0] = 2 A [1] = 5678; A [2] = 1234; B [0] = 3 B [1] = 9012; B [2] = 5678; B [3] = 1234; */P = 0; for (I = 0; I <= K; I ++) C [I] = 0; C [I] = 0; // A [1], A [2]..., A [m] // B [1] // C [1], C [2],..., c [m] For (j = 1; j <= N; j ++) {S = J; If (B [J]) for (I = 1; I <= m; I ++, s ++) {C [s] + = A [I] * B [J]; If (C [s]> = Base) {c [S + 1] + = C [s]/base; C [s] % = base ;}}for (I = 1; I <= K; I ++) if (C [I]> = base) {C [I + 1] + = C [I]/base; C [I] % = base ;} if (C [k + 1]) K ++; C [0] = K;} // C = A/B k = A % bint Div (int *, int B, int * c) {int I, j, S, T; int K = 0, Q; // long K = 0, Q; base >=100000 if (COMP (A, zero) = 0) {copy (C, zero); Return 0 ;}for (I = A [0]; i> = 1; I --) {q = K * base + A [I]; C [I] = Q/B; k = Q % B ;} if (C [A [0] = 0) {C [0] = A [0]-1;} else C [0] = A [0]; return K;} // big integer except Big integer c = A/B, r = A % bvoid Div (int * a, int * B, int * C, int * r) {int M = A [0], n = B [0], I, J, K, S, T, P, Q, X, Y; int base1000 = base * 1000, base100 = base * 100, base10 = base * 10; // M digits except n digits = k = m-n + 1 digits if (COMP (a, B) <0) {copy (R, a); copy (C, zero); return;} k = m-n + 1; for (S = K; S> = 0; s --) C [s] = 0; for (S = K, t = m; S> = 1, t> = N;) {P = comp (a, B, T ); // A [T]-B [N] // If (p <0 & t = N) break; If (P = 0) {C [s] ++; For (j = T; j> = T-n + 1; j --) A [J] = 0; t-= N; If (T <n) break; s-= N;} If (p> 1) {C [s] ++; for (j = t, q = N; j> = T-n + 1; j --, Q --) A [J]-= B [Q]; for (j = T-n + 1; j <= T; j ++) {if (a [J] <0) {A [J] + = base; A [J + 1] --;} //} For (j = T; j> = T-n + 1; j --) if (a [J]) break; If (j <n) break; s-= T-J; T = J;} If (p <0 & T> N) {A [T-1] + = A [T] * base; A [T] = 0; t --; S --;} If (P = 1) // 9999 1234/1999 9999 x = 4 {x = A [T]/(B [N] + 1 ); c [s] + = x; // x <4999 for (j = t, q = N; q> = 1; q --, j --) A [J]-= x * B [Q]; for (j = T-n + 1; j <= T; j ++) {While (A [J] <=-base1000) {A [J] + = base1000; A [J + 1]-= 1000 ;} while (A [J] <=-base100) {A [J] + = base100; A [J + 1]-= 100 ;} while (A [J] <=-base10) {A [J] + = base10; A [J + 1]-= 10;} while (A [J] <0) {A [J] + = base; A [J + 1] -- ;}}} if (C [k] = 0) C [0] = K-1; else C [0] = K; for (I = m; I> = 1; I --) if (a [I]) break; if (I = 0) copy (R, zero); else {R [0] = I; while (I >= 1) {R [I] = A [I]; I --;}}} // d = gcd (a, B) void gcd (int * a, int * B, int * D) {int P, X [10000], Y [10000], c [10000]; P = comp (a, B); If (p <0) {copy (X, B); copy (Y, );} else {copy (X, A); copy (Y, B);} // x> = B if (COMP (Y, zero) = 0) {copy (D, x); return;} copy (D, Y); While (1) {Div (x, D, C, Y); // C = x/d; y = x % d if (COMP (Y, zero) = 0) {return;} If (COMP (Y, one) = 0) {copy (D, one); return;} copy (x, D); copy (D, Y );}}