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(1)另一种方法,有问题?
/*poj上WA因为最近在学计算机算法分析与设计的递归与分治策略,书中有提到Fibonacci数列,我想到了Fibonacci数列的非递归方式。但是不知道为什么,n<45时,输出和ac代码的输出是一样的,WA,下文有测试代码和输出截图。*/#include#include intf(int n) { int a; a = 1.0 / sqrt(5.0) * (pow((1 + sqrt(5.0)) / 2, n + 2) - pow((1 - sqrt(5.0)) / 2, n + 2)); return a;}intmain() { int n, k, i, ans; scanf("%d", &n); for( i = 1; i <= n; i++ ) { scanf("%d", &k); ans = f(k); printf("Scenario #%d:\n%d\n\n", i, ans); } return 0;}
(2)测试代码
#include#include intf(int n) { int a; a = 1.0 / sqrt(5.0) * (pow((1 + sqrt(5.0)) / 2, n + 2) - pow((1 - sqrt(5.0)) / 2, n + 2)); return a;}intmain() { int i, mans; int ans[50] = { 1, 2, 3}; for( i = 3; i < 45; i++ ) { ans[i] = ans[i - 1] + ans[i - 2]; } for( i = 1; i < 45; i++ ) { mans = f(i); printf("i = %d, %d =? %d\n", i, ans[i], mans); } return 0;}
(3)难道是因为这样?
/*我请教老师,他说让我取n = 100试一下,谢谢老师。开始我用int直接做了,结果 n = 45 时,就溢出了。后面我又改用long long int试了一次,发现 n = 70时,用数组打表(ac代码)和用非递归的数学公式的结果就开始不一样了。(环境:win10专业版,codeblocks16.01)*/
(4)int测试代码
#include#include intf(int n) { int a; a = 1.0 / sqrt(5.0) * (pow((1 + sqrt(5.0)) / 2, n + 2) - pow((1 - sqrt(5.0)) / 2, n + 2)); return a;}intmain() { int i, mans; int ans[100] = { 1, 2, 3}; for( i = 3; i < 100; i++ ) { ans[i] = ans[i - 1] + ans[i - 2]; } for( i = 1; i < 100; i++ ) { mans = f(i); printf("i = %d, %d =? %d\n", i, ans[i], mans); } return 0;}
(5)long long int测试代码
#include#include long longf(int n) { long long a; a = 1.0 / sqrt(5.0) * (pow((1 + sqrt(5.0)) / 2, n + 2) - pow((1 - sqrt(5.0)) / 2, n + 2)); return a;}intmain() { int i; long long mans; long long ans[105] = { 1, 2, 3}; for( i = 3; i < 100; i++ ) { ans[i] = ans[i - 1] + ans[i - 2]; } for( i = 1; i < 100; i++ ) { mans = f(i); printf("i = %d, %I64d =? %I64d\n", i, ans[i], mans); } return 0;}
(6)可是题目是这样说的
/*The first line contains the number of scenarios.For each scenario, you are given a single positive integer less than 45 on a line by itself.不是说好less than 45的吗?*/