SGU 116 Index of superprime

SGU_116

    一开始用回溯写了一下，发现效率还是很高的，但需要注意不是当前选择的越大就一定越好。

    后来看了别人的题解提到了dp的解法，于是便又用dp写了一下，用f[i]表示i最少可以分解为几个超级素数之和，同时用fa[]记录下决策的过程即可。

//回溯
#include<stdio.h>
#include<string.h>
#define MAXD 10010
#define INF 0x3f3f3f3f
int N, isprime[MAXD], prime[MAXD], p, sprime[MAXD], sp, a[MAXD], ans[MAXD], num;
void prepare()
{
    int i, j, k;
    memset(isprime, -1, sizeof(isprime));
    p = 0;
    for(i = 2; i <= 10000; i ++)
        if(isprime[i])
        {
            prime[++ p] = i;
            for(j = i * i; j <= 10000; j += i)
                isprime[j] = 0;
        }
    sp = 0;
    for(i = 2; i <= p; i ++)
        if(isprime[i])
            sprime[++ sp] = prime[i];
}
void dfs(int t, int cur, int s)
{
    int i, j;
    if(cur >= num)
        return ;
    if(s == 0)
    {
        num = cur;
        for(i = 0; i < cur; i ++)
            ans[i] = a[i];
        return ;
    }
    for(i = t; i > 0; i --)
        if(s >= sprime[i])
        {
            a[cur] = sprime[i];
            dfs(i, cur + 1, s - sprime[i]);
        }
}
void solve()
{
    int i, j ,k;
    num = 0x3f3f3f3f;
    dfs(sp, 0, N);
    if(num == INF)
        printf("0\n");
    else
    {
        printf("%d\n", num);
        printf("%d", ans[0]);
        for(i = 1; i < num; i ++)
            printf(" %d", ans[i]);
        printf("\n");
    }
}
int main()
{
    int i, j, k;
    prepare();
    while(scanf("%d", &N) == 1)
    {
        solve();
    }
    return 0;
}

//dp
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define MAXD 10010
#define INF 0x3f3f3f3f
int N, isprime[MAXD], prime[MAXD], p, sprime[MAXD], sp, a[MAXD], ans[MAXD], num, f[MAXD], fa[MAXD];
int cmp(const void *_p, const void *_q)
{
    int *p = (int *)_p;
    int *q = (int *)_q;
    return *q - *p;
}
void prepare()
{
    int i, j, k;
    memset(isprime, -1, sizeof(isprime));
    p = 0;
    for(i = 2; i <= 10000; i ++)
        if(isprime[i])
        {
            prime[++ p] = i;
            for(j = i * i; j <= 10000; j += i)
                isprime[j] = 0;
        }
    sp = 0;
    for(i = 2; i <= p; i ++)
        if(isprime[i])
            sprime[++ sp] = prime[i];
    memset(f, 0x3f, sizeof(f));
    f[0] = 0;
    for(i = 1; i <= 10000; i ++)
        for(j = 1; j <= sp && sprime[j] <= i; j ++)
            if(f[i - sprime[j]] + 1 < f[i])
            {
                f[i] = f[i - sprime[j]] + 1;
                fa[i] = i - sprime[j];
            }
}
void solve()
{
    int i, j ,k;
    num = 0x3f3f3f3f;
    if(f[N] == INF)
        printf("0\n");
    else
    {
        for(i = N, k = 0; i != 0; i = fa[i])
            ans[k ++] = i - fa[i];
        qsort(ans, k, sizeof(ans[0]), cmp);
        printf("%d\n", k);
        printf("%d", ans[0]);
        for(i = 1; i < k; i ++)
            printf(" %d", ans[i]);
        printf("\n");
    }
}
int main()
{
    int i, j, k;
    prepare();
    while(scanf("%d", &N) == 1)
    {
        solve();
    }
    return 0;
}