在进入问题描述之前，请注意数据类型＆＃39; factrep＆＃39;仅仅是vector<int>的typedef。
问题
鉴于两个factrep个对象f1＆amp; f2，作为参数传递到factrep函数mult，我希望返回的factrep对象result能够表示相乘的向量。
另请注意，对矢量执行的实际算术运算＆＃39;单独的元素是加法：这是由于向量以素数因子指数的形式存储（元素0 =素数2到存储的任何东西的幂，元素1 =素数3到存储的任何幂的幂）。因此，添加元素应该是正确的，但我的结果并不一致。我将把结果发布在附录中。
这是相关功能
factrep mult(factrep f1, factrep f2)
{
factrep a;
for(unsigned int i=0; i <f1.size() && f2.size(); i++) a.push_back(0);
for(int i = 0; i < a.size(); i++)
{
a[i] += f1[i]+f2[i];
}
return a;
}
这是完整的代码，包括调试控制台打印
#include "facth_new.h"
#include <iostream>
#include <cmath>
#include <climits>
#include <vector>
using namespace std;
std::vector<int> eratosthenes(int n) {
std::vector<int> result = std::vector<int>();
if (n < 2) {
return result;
}
// initialize the vector
std::vector<bool> input(n + 1, true);
// calculate the upper limit as the square root of
// of N. all composite numbers <= N must have a
// factor <= sqrt(N)
int sqrtN = (int)sqrt(n);
// iterate from 2 up to the square root of N
for (int i = 2; i <= sqrtN; i ++) {
if (! input[i]) {
// i is a proven composite number,
// all its multiples have been
// marked not prime by all its prime factors by now.
continue;
}
// as an optimization, all multiples *less* than
// the square of i are already marked, (they have
// another prime factor less than i), so we can start
// from the square which is the smallest composite
// not yet marked
for (int j = i * i; j <= n; j += i) {
input[j] = false;
}
}
// as n >= 2, then add 2 here
result.push_back(2);
// and check only odd numbers here,
// no other even number can be set
for (int i = 3; i <= n; i += 2) {
if (input[i]) {
result.push_back(i);
}
}
return result;
}
factrep primfact(int n){
factrep a;
int m; // still to factorize number
m=n;
// continue until nothing to factorize
for(int i = 0; m != 1; i++)
{
a.push_back(0);
while(m % primes[i] == 0){
m=m/primes[i];
a.at(i)++;
}
}
return a;
}
factrep mult(factrep f1, factrep f2)
{
factrep a;
for(unsigned int i=0; i <f1.size() && f2.size(); i++) a.push_back(0);
for(int i = 0; i < a.size(); i++)
{
a[i] += f1[i]+f2[i];
}
return a;
}
factrep div(factrep f1, factrep f2)
{
factrep result;
for(int i=0; i<f1.size(); i++){
result.push_back(f1[i]-f2[i]);
}
return result;
}
double getval(factrep f)
{
double result = 1;
for(unsigned int i = 0; i < f.size(); i++)
{
result *= pow(primes[i],f[i]);
}
return result;
}
vector<int> primes;
int main() {
primes = eratosthenes(71);
cout << "The prime numbers are:\n";
for(unsigned i = 0; i != primes.size(); i++)
cout << primes[i] << '\n';
int num[] = {1, 17, 54, 10, 36, 63, 20, 25};
factrep f[8];
for(int i = 0; i != 8; i++) {
f[i] = primfact(num[i]);
}
for(int i = 0; i != 8; i++) {
cout << '\n' << num[i] << " is factorized as:\nPrime Exponent\n";
bool agree = true;
for(unsigned j = f[i].size(); j < f[i].size(); j++) {
if((f[i])[j] != 0) {
agree = false;
}
}
for(unsigned j = 0; j != min(f[i].size(), f[i].size()); j++) {
if((f[i])[j] != 0) {
cout << primes[j] << " " << (f[i])[j] << '\n';
}
}
if(!agree) {
for(unsigned j = f[i].size(); j != f[i].size(); j++) {
if((f[i])[j] != 0) {
cout << primes[j] << " " << (f[i])[j] << '\n';
}
}
}
}
for(int i = 0; i != 8; i++) {
cout << "getval of " << num[i] << " is " << getval(f[i]) << "\n\n";
}
for(int i = 0; i != 8; i++) {
for(int j = i; j != 8; j++) {
factrep res = mult(f[i], f[j]);
cout << num[i] << " multiplied by " << num[j] << " is " << getval(res) << "\n";
}
}
for(int i = 0; i != 8; i++) {
for(int j = i; j != 8; j++) {
factrep res = div(f[i], f[j]);
cout << num[i] << " divided by " << num[j] << " is " << getval(res) << "\n";
}
}
}
最后，我的调试控制台打印出在乘法部分发生的错误：
The prime numbers are:
2
3
5
7
11
13
17
19
23
29
31
37
41
43
47
53
59
61
67
71
1 is factorized as:
Prime Exponent
17 is factorized as:
Prime Exponent
17 1
54 is factorized as:
Prime Exponent
2 1
3 3
10 is factorized as:
Prime Exponent
2 1
5 1
36 is factorized as:
Prime Exponent
2 2
3 2
63 is factorized as:
Prime Exponent
3 2
7 1
20 is factorized as:
Prime Exponent
2 2
5 1
25 is factorized as:
Prime Exponent
5 2
getval of 1 is 1
getval of 17 is 17
getval of 54 is 54
getval of 10 is 10
getval of 36 is 36
getval of 63 is 63
getval of 20 is 20
getval of 25 is 25
1 multiplied by 1 is 1
1 multiplied by 17 is 1
1 multiplied by 54 is 1
1 multiplied by 10 is 1
1 multiplied by 36 is 1
1 multiplied by 63 is 1
1 multiplied by 20 is 1
1 multiplied by 25 is 1
17 multiplied by 17 is 289
17 multiplied by 54 is 9.12788e+172
17 multiplied by 10 is 1.69035e+172
17 multiplied by 36 is 3.04263e+173
17 multiplied by 63 is 1.06492e+173
17 multiplied by 20 is 2.36649e+173
17 multiplied by 25 is 2.95811e+173
54 multiplied by 54 is 2916
54 multiplied by 10 is 108
54 multiplied by 36 is 1944
54 multiplied by 63 is 486
54 multiplied by 20 is 216
54 multiplied by 25 is 54
10 multiplied by 10 is 100
10 multiplied by 36 is 9000
10 multiplied by 63 is 90
10 multiplied by 20 is 200
10 multiplied by 25 is 250
36 multiplied by 36 is 1296
36 multiplied by 63 is 324
36 multiplied by 20 is 144
36 multiplied by 25 is 36
63 multiplied by 63 is 3969
63 multiplied by 20 is 61740
63 multiplied by 25 is 540225
20 multiplied by 20 is 400
20 multiplied by 25 is 500
25 multiplied by 25 is 625
1 divided by 1 is 1
1 divided by 17 is 1
1 divided by 54 is 1
1 divided by 10 is 1
1 divided by 36 is 1
1 divided by 63 is 1
1 divided by 20 is 1
1 divided by 25 is 1
17 divided by 17 is 1
17 divided by 54 is 0
17 divided by 10 is 0
17 divided by 36 is 0
17 divided by 63 is 0
17 divided by 20 is 0
17 divided by 25 is 0
54 divided by 54 is 1
54 divided by 10 is 27
54 divided by 36 is 1.5
54 divided by 63 is 6
54 divided by 20 is 13.5
54 divided by 25 is 54
10 divided by 10 is 1
10 divided by 36 is 0.277778
10 divided by 63 is 1.11111
10 divided by 20 is 0.5
10 divided by 25 is 0.4
36 divided by 36 is 1
36 divided by 63 is 4
36 divided by 20 is 9
36 divided by 25 is 36
63 divided by 63 is 1
63 divided by 20 is 3.15
63 divided by 25 is 0.36
20 divided by 20 is 1
20 divided by 25 is 0.8
25 divided by 25 is 1
答案 0 :(得分：1)
条件i <f1.size() && f2.size()
应该写成i <f1.size() && i < f2.size()
答案 1 :(得分：1)
你的for循环条件：
for(unsigned int i=0; i <f1.size() && f2.size(); i++)
应该是：
for(unsigned int i=0; i<f1.size() && i<f2.size(); i++)
这意味着我从0循环到min {f1.size（），f2.size（）}
答案 2 :(得分：1)
正如其他人所说，问题在于你的循环状况。除此之外，我还提到了一些你可以改变的东西，以使你的代码更符合逻辑。
您有以下两个循环
for(unsigned j = f[i].size(); j < f[i].size(); j++) {
if((f[i])[j] != 0) {
agree = false;
}
}
for(unsigned j = f[i].size(); j != f[i].size(); j++) {
if((f[i])[j] != 0) {
cout << primes[j] << " " << (f[i])[j] << '\n';
}
}
在这两个中你已经用f [i] .size（）初始化j，然后设置条件以结束循环，如果j == f [i] .size（）。由于j已经是f [i] .size（），这将在第一次迭代后中断。在这种情况下，你根本不需要循环，而只需要你的if语句。