C++11处女题,时间竟然用了790ms,无语啊,咋会这么慢,感觉算法差不多啊。。算了下次用C++试下,妹的新特性的for循环和JAVA好像,用起来爽,不是是不是因为C++11的原因这么慢。。题目的话没什么好说的,简单DP,类似完全背包,注意题目是递减输出,我开始看到递增了,WA了一次
#include<iostream>
#include<vector>
#include<cstring>
using namespace std;
vector<int> primes(int max)
{
vector<int> res { 2 };
for (int i = 3; i <= max; i += 2)
{
bool is_prime = true;
for (int p : res)
{
if (i % p == 0)
{
is_prime = false;
break;
}
if (p * p > i)
break;
}
if (is_prime)
res.push_back(i);
}
return res;
}
int main()
{
vector<int> res = primes(10000);
int dp[10001], path[10001];
memset(dp, 1, sizeof(dp));
memset(path, 0, sizeof(path));
dp[0] = 0;
for (int prime : res)
for (int i = prime; i <= 10000; i++)
{
if (dp[i - prime] + 1 < dp[i])
{
dp[i] = dp[i - prime] + 1;
path[i] = prime;
}
}
int num;
while (cin >> num)
{
if (num == 0 || num == 1)
{
cout << 0 << endl;
continue;
}
cout << dp[num] << endl;
while (num)
{
if (path[num] == num)
cout << num;
else
cout << path[num] << " ";
num -= path[num];
}
cout << endl;
}
return 0;
}
我去,不信邪把原来的C++11的特性全去掉了,改用旧C++重写了一遍,时间就提升到20ms了,无语。。新特性难道就这么费时间么,老实把C++0x把编译选项去掉了,默默地用老C++写吧。。唉,不与时俱进啊
#include<cstdio>
#include<vector>
#include<cstring>
using namespace std;
vector<int> primes(int max)
{
vector<int> res;
res.push_back(2);
for (int i = 3; i <= max; i += 2)
{
bool is_prime = true;
int p;
for (vector<int>::iterator it = res.begin(); it != res.end(); it++)
{
p = *it;
if (i % p == 0)
{
is_prime = false;
break;
}
if (p * p > i)
break;
}
if (is_prime)
res.push_back(i);
}
return res;
}
int main()
{
vector<int> res = primes(10000);
int dp[10001], path[10001];
memset(dp, 1, sizeof(dp));
memset(path, 0, sizeof(path));
dp[0] = 0;
int prime;
for (vector<int>::iterator it = res.begin(); it != res.end(); it++)
{
prime = *it;
for (int i = prime; i <= 10000; i++)
{
if (dp[i - prime] + 1 < dp[i])
{
dp[i] = dp[i - prime] + 1;
path[i] = prime;
}
}
}
int num;
while (scanf("%d", &num) != EOF)
{
if (num == 0 || num == 1)
{
puts("0");
continue;
}
printf("%d\n", dp[num]);
while (num)
{
if (path[num] == num)
printf("%d", num);
else
printf("%d ", path[num]);
num -= path[num];
}
putchar('\n');
}
return 0;
}