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¿Cuál es la diferencia entre el código javascript y python Fibonacci aquí?

Actualmente estoy usando fórmulas de multiplicación de matrices para crear algoritmos para secuencias de Fibonacci.

Hice mi código JavaScript basado en el siguiente código python. Sin embargo, se confirmó que se emitió otro valor [case fib(150)]. Creo que el mod % está mal, pero ¿cómo puedo solucionar este problema?

codigo javascript

 const n = parseInt(prompt("Number")); const mod = 1000000007; const fib = () => { const zero = [ [1, 1], [1, 0], ]; const base = [ [1], [1] ]; const power = (a, num) => { if (num === 1) return a; else if (num % 2 != 0) return multi(power(a, num - 1), a); else return power(multi(a, a), parseInt(num / 2)); }; const multi = (a, b) => { const temp = Array.from(Array(2), () => Array(b[0].length).fill(0)); for (let i = 0; i < 2; i++) { for (let j = 0; j < b[0].length; j++) { let sum_num = 0; for (let k = 0; k < 2; k++) { sum_num += a[i][k] * b[k][j]; } temp[i][j] = sum_num % mod; } } return temp; }; return multi(power(zero, n - 2), base)[0][0]; }; if (n === 0) { console.log(0); } else if (n < 3) { console.log(1); } else { console.log(fib()); }

código pitón

 import sys input = sys.stdin.readline MOD = 1000000007 adj=[[1,1],[1,0]] start=[[1],[1]] N = int(input()) def power(adj,n): if n == 1: return adj elif n % 2: return multi(power(adj,n-1), adj) else: return power(multi(adj,adj), n//2) def multi(a,b): temp=[[0]*len(b[0]) for _ in range(2)] for i in range(2): for j in range(len(b[0])): sum_n = 0 for k in range(2): sum_n += a[i][k]*b[k][j] temp[i][j]= sum_n % MOD return temp if N < 3: print(1) else: print(multi(power(adj,N-2),start)[0][0])
about 4 years ago · Juan Pablo Isaza
1 answers
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0

Lo resolví usando bigint. Gracias por tu ayuda. (mplungjan, kelly bundy)

 const n = BigInt(prompt('Number')); const mod = 1000000007n; const fib = () => { const zero = [ [1n, 1n], [1n, 0n], ]; const base = [[1n], [1n]]; const power = (a, num) => { if (num === 1n) return a; else if (num % 2n != 0) return multi(power(a, num - 1n), a); else return power(multi(a, a), BigInt(num / 2n)); }; const multi = (a, b) => { const temp = Array.from(Array(2), () => Array(b[0].length).fill(0)); for (let i = 0; i < 2; i++) { for (let j = 0; j < b[0].length; j++) { let sum_num = 0n; for (let k = 0; k < 2; k++) { sum_num += a[i][k] * b[k][j]; } temp[i][j] = sum_num % mod; } } return temp; }; return multi(power(zero, n - 2n), base)[0][0]; }; if (n === 0) { console.log(0); } else if (n < 3) { console.log(1); } else { console.log(parseInt(fib())); }
about 4 years ago · Juan Pablo Isaza Report
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