C++类实现大数计算的问题

1个回答

**题目描述** 幂运算是常见的数学运算之一，其原理是用同一个数相乘多次，但是有的时候当幂指数特别大的时候，这样的运算就太浪费时间。请大家学会在幂中加特技，让幂运算的效率提高到可以接受的程度。 **输入：** 第一个行一个整数T，表示有T组数据 每组数据，输入x，y 求x的y次幂 （2≤ x ,y≤10^9） **输出：** 每组数据输出一个整数，表示幂运算对1000000007取模后的结果 **样例输入：** 2 2 4 2 100000000 **样例输出：** 16 494499948 我的代码总是超时，求好方法！！谢谢！！

C语言的函数算法去解决一个矩阵的问题，涉及到大数的计算怎么实现的？

Problem Description In mathematics, the Fibonacci numbers are a sequence of numbers named after Leonardo of Pisa, known as Fibonacci (a contraction of filius Bonaccio, "son of Bonaccio"). Fibonacci's 1202 book Liber Abaci introduced the sequence to Western European mathematics, although the sequence had been previously described in Indian mathematics. The first number of the sequence is 0, the second number is 1, and each subsequent number is equal to the sum of the previous two numbers of the sequence itself, yielding the sequence 0, 1, 1, 2, 3, 5, 8, etc. In mathematical terms, it is defined by the following recurrence relation: That is, after two starting values, each number is the sum of the two preceding numbers. The first Fibonacci numbers (sequence A000045 in OEIS), also denoted as F[n]; F[n] can be calculate exactly by the following two expressions: A Fibonacci spiral created by drawing arcs connecting the opposite corners of squares in the Fibonacci tiling; this one uses squares of sizes 1, 1, 2, 3, 5, 8, 13, 21, and 34; So you can see how interesting the Fibonacci number is. Now AekdyCoin denote a function G(n) Now your task is quite easy, just help AekdyCoin to calculate the value of G (n) mod C Input The input consists of T test cases. The number of test cases (T is given in the first line of the input. Each test case begins with a line containing A, B, N, C (10<=A, B<2^64, 2<=N<2^64, 1<=C<=300) Output For each test case, print a line containing the test case number( beginning with 1) followed by a integer which is the value of G(N) mod C Sample Input 1 17 18446744073709551615 1998 139 Sample Output Case 1: 120

Description Computing the exact number of ways that N things can be taken M at a time can be a great challenge when N and/or M become very large. Challenges are the stuff of contests. Therefore, you are to make just such a computation given the following: GIVEN: 5 <= N <= 100; 5 <= M <= 100; M <= N Compute the EXACT value of: C = N! / (N-M)!M! You may assume that the final value of C will fit in a 32-bit Pascal LongInt or a C long. For the record, the exact value of 100! is: 93,326,215,443,944,152,681,699,238,856,266,700,490,715,968,264,381,621, 468,592,963,895,217,599,993,229,915,608,941,463,976,156,518,286,253, 697,920,827,223,758,251,185,210,916,864,000,000,000,000,000,000,000,000 Input The input to this program will be one or more lines each containing zero or more leading spaces, a value for N, one or more spaces, and a value for M. The last line of the input file will contain a dummy N, M pair with both values equal to zero. Your program should terminate when this line is read. Output The output from this program should be in the form: N things taken M at a time is C exactly. Sample Input 100 6 20 5 18 6 0 0 Sample Output 100 things taken 6 at a time is 1192052400 exactly. 20 things taken 5 at a time is 15504 exactly. 18 things taken 6 at a time is 18564 exactly.

Problem Description Xiao Jun likes math and he has a serious math question for you to finish. Define F[x] to the xor sum of all digits of x under the decimal system,for example F(1234) = 1 xor 2 xor 3 xor 4 = 4. Two numbers a,b(a≤b) are given,figure out the answer of F[a] + F[a+1] + F[a+2]+…+ F[b−2] + F[b−1] + F[b] doing a modulo 109+7. Input The first line of the input is a single integer T(T<26), indicating the number of testcases. Then T testcases follow.In each testcase print three lines : The first line contains one integers a. The second line contains one integers b. 1≤|a|,|b|≤100001,|a| means the length of a. Output For each test case, output one line "Case #x: y", where x is the case number (starting from 1) and y is the answer. Sample Input 4 0 1 2 2 1 10 9999 99999 Sample Output Case #1: 1 Case #2: 2 Case #3: 46 Case #4: 649032

Problem Description FSF has programmed a game. In this game, players need to divide a rectangle into several same squares. The length and width of rectangles are integer, and of course the side length of squares are integer. After division, players can get some coins. If players successfully divide a AxB rectangle(length: A, width: B) into KxK squares(side length: K), they can get A*B/ gcd(A/K,B/K) gold coins. In a level, you can’t get coins twice with same method. (For example, You can get 6 coins from 2x2(A=2,B=2) rectangle. When K=1, A*B/gcd(A/K,B/K)=2; When K=2, A*B/gcd(A/K,B/K)=4; 2+4=6; ) There are N*(N+1)/2 levels in this game, and every level is an unique rectangle. (1x1 , 2x1, 2x2, 3x1, ..., Nx(N-1), NxN) FSF has played this game for a long time, and he finally gets all the coins in the game. Unfortunately ,he uses an UNSIGNED 32-BIT INTEGER variable to count the number of coins. This variable may overflow. We want to know what the variable will be. (In other words, the number of coins mod 2^32) Input There are multiply test cases. The first line contains an integer T(T<=500000), the number of test cases Each of the next T lines contain an integer N(N<=500000). Output Output a single line for each test case. For each test case, you should output "Case #C: ". first, where C indicates the case number and counts from 1. Then output the answer, the value of that UNSIGNED 32-BIT INTEGER variable. Sample Input 3 1 3 100 Sample Output Case #1: 1 Case #2: 30 Case #3: 15662489

RT 比如说，2的10000000次方，我用double倒是可以算，但是如何去模呢。。。 2的10000000次方对1234567取模。。。 谢谢大神们！

Problem Description We all know the definition of Fibonacci series: fib[i]=fib[i-1]+fib[i-2],fib[1]=1,fib[2]=1.And we define another series P associated with the Fibonacci series: P[i]=fib[4*i-1].Now we will give several queries about P:give two integers L,R, and calculate ∑P[i](L <= i <= R). Input There is only one test case. The first line contains single integer Q – the number of queries. (Q<=10^4) Each line next will contain two integer L, R. (1<=L<=R<=10^12) Output For each query output one line. Due to the final answer would be so large, please output the answer mod 1000000007. Sample Input 2 1 300 2 400 Sample Output 838985007 352105429

Problem Description In many applications very large integers numbers are required. Some of these applications are using keys for secure transmission of data, encryption, etc. In this problem you are given a number, you have to determine the number of digits in the factorial of the number. Input Input consists of several lines of integer numbers. The first line contains an integer n, which is the number of cases to be tested, followed by n lines, one integer 1 ≤ n ≤ 107 on each line. Output The output contains the number of digits in the factorial of the integers appearing in the input. Sample Input 2 10 20 Sample Output 7 19

Problem Description Everyone knows that the letter “E” is the most frequent letter in the English language. In fact, there are one hundred sixteen E’s on this very page ... no, make that one hundred twenty one. Indeed, when spelling out integers it is interesting to see which ones do NOT use the letter “E”. For example 6030 (six thousand thirty) doesn’t. Nor does 4002064 (four million two thousand sixty four). It turns out that 6030 is the 64th positive integer that does not use an “E” when spelled out and 4002064 is the 838th such number. Your job is to find the n-th such number. Note: 1,001,001,001,001,001,001,001,001,000 is “one octillion, one septillion, one sextillion, one quintil-lion, one quadrillion, one trillion, one billion, one million, one thousand”. (Whew!) Input The input file will consist of multiple test cases. Each input case will consist of one positive integer n (less than 231) on a line. A 0 indicates end-of-input. (There will be no commas in the input.) Output For each input n you will print, with appropriate commas, the n-th positive integer whose spelling does not use an “E”. You may assume that all answers are less than 1028. Sample Input 1 10 838 0 Sample Output 2 44 4,002,064

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abs函数求取绝对值，int函数取整，怎么将两个函数结合起来对小数点后的数据实现取整运算？

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int main() { int a, b, c, d, e, f, g, h, i, j; scanf_s("%d,%d,%d,%d,%d,%d,%d,%d,%d,%d", &a, &b, &c, &d, &e, &f, &g, &h, &i, &j); if (a > b && a > c && a> d && a > e && a > f && a > g && a > h && a > i && a > j) printf("%d\n", a); else if (b > a && c > c && b > d && b > e && b > f && b > g && b > h && b > i && b > j) printf("%d\n", b); else if (c > a && c > b && c > d && c > e && c > f && c > g && c > h && c > i && c > j) printf("%d\n", b); else if (d > a && d > b && d > c && d > e && d > f && d > g && d > h && d > i && d > j) printf("%d\n", b); else if (e > a && e > b && e > c && e > d && e > f && e > g && e > h && e > i && e > j) printf("%d\n", b); else if (f > a && f > b && f > c && f > d && f > e && e > g && e > h && e > i && e > j) printf("%d\n", b); else if (g > a && g > b && g > c && g > c && g > e && g > f && g > h && g > i && g > j) printf("%d\n", b); else if (h > a && h > b && h > c && h > d && h > e && h > f && h > g && h > i && h > j) printf("%d\n", b); else if (i > a && i > b && i > c && i > d && i > e && i > f && i > g && i > h && i > j) printf("%d\n", b); else printf("%d\n", b); return 0; } 主要是思考算法

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tb={} tb[1]={"A1","A2","A3"} tb[2]={"B1","B2","B3","B4"} tb[3]={"C1","C2","C3","C4","C5","C6"} d={} for a=1,#tb[1] do for b=1,#tb[2] do for c=1,#tb[3] do table.insert(d,{tb[1][a],tb[2][b],tb[3][c]}) end end end 这是已知数组下标的情况下写的3层for循环能达到要求,那么要是下标是5 或者6呢,如何做到自适应?

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