P1832 A+B Problem（再升级）

题目背景

·题目名称是吸引你点进来的

·实际上该题还是很水的

题目描述

·1+1=? 显然是2

·a+b=? 1001回看不谢

·哥德巴赫猜想 似乎已呈泛滥趋势

·以上纯属个人吐槽

·给定一个正整数n，求将其分解成若干个素数之和的方案总数。

输入输出格式

输入格式：

一行：一个正整数n

输出格式：

一行：一个整数表示方案总数

输入输出样例

输入样例#1：

7

输出样例#1：

3

说明

【样例解释】

7=7 7=2+5

7=2+2+3

【福利数据】

【输入】 20

【输出】 26

【数据范围及约定】

对于30%的数据 1<=n<=10

对于100%的数据，1<=n<=10^3

埃拉托斯提尼筛法打素数表。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>

inline long long read()
{
	long long x = 0;char ch = getchar();char c = ch;
	while(ch > ‘9‘ || ch < ‘0‘)c = ch,ch = getchar();
	while(ch <= ‘9‘ && ch >= ‘0‘)x = x * 10 + ch - ‘0‘,ch = getchar();
	if(c == ‘-‘)return -1 * x;
	return x;
}

const int INF = 0x3f3f3f3f;
const int MAXN = 0x4f4f4f4f;

bool b[MAXN];

int main()
{
	long long n = read();
	for(int i = 2;i*i <= n;i ++)
	{
		if(!b[i])
		{
			for(int j = i*2;j <= n;j += i)
			{
				b[j] = true;
			}
		}
	}
	long long ans = 0;
	for(int i = 2;i <= n;i ++)
	{
		if(!b[i])
		{
			printf("%d,", i);
			ans ++;
		}
	}
	printf("\n%d", ans);
	return 0;
}

素数输出到文件，复制粘贴过来直接用。

使用背包统计次数，是一个完全背包的小变形（如果真正理解了完全背包，直接看代码就能懂）。初始化为f[0] = 1，因为f[v[i]] = 1，转移时f[v[i]] = f[v[i] - v[i]]。

代码如下。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>

inline int read()
{
	int x = 0;char ch = getchar();char c = ch;
	while(ch > ‘9‘ || ch < ‘0‘)c = ch,ch = getchar();
	while(ch <= ‘9‘ && ch >= ‘0‘)x = x * 10 + ch - ‘0‘,ch = getchar();
	if(c == ‘-‘)return -1 * x;
	return x;
}

const int INF = 0x3f3f3f3f;
int prime[170] = {0,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997};
int n;
long long f[1010];
int main()
{
	n = read();
	f[0] = 1;
	for(int i = 1;prime[i] <= n && i <= 168;i ++)
	{
		for(int j = prime[i];j <= n;j ++)
		{
			f[j] += f[j - prime[i]];
		}
	}
	std::cout<<f[n];
	return 0;
}