SPOJ CZ_PROB1 - Summing to a Square Prime

题目链接：http://www.spoj.com/problems/CZ_PROB1/

题目大意：Sp2 是所有素数中能够被分解为两个数的平方和的素数的集合。P(a,b)指的是在a的划分中，最大元素不超过b的划分的个数（就是那个经典整数划分动态规划题）。给你个n,k，问P(sp2[n],k)是多少。

• 0<T<501
• 0<n<501
• 1<SP2(n)<7994
• 0<k<4

解题思路：既然题中已经给出sp2集合的限制，那么sp2直接打表，按照经典动态规划算法计算p(a,b)然后再打表，然后。。。就没有然后了

转移方程：

if(j > i)　　p[i][j] = p[i][i];
else if(j == i)　　p[i][j] = p[i][j - 1] + 1;
else　　p[i][j] = p[i][j - 1] + p[i - j][j];

初始化的时候 p[1][1] = p[1][2] = p[1][3] = p[1][4] = 1，其余0即可。

代码：

const int maxn = 1e4 + 5;
bool vis[maxn];
vector<int> vec;
int n, k;
int p[maxn][10];

bool isprime(int x){
    if(x == 2 || x == 3) return true;
    if(x == 1 || x % 2 == 0) return false;
    for(int i = 2; i * i <= x; i++)
        if(x % i == 0) return false;
    return true;
}
void dowork(){
    memset(vis, 0, sizeof(vis));
    for(int i = 1; i <= 90; i++){
        for(int j = 1; j <= 90; j++){
            int x = i * i + j * j;
            if(isprime(x) && x < maxn && vis[x] == 0) {
                vec.push_back(x);
                vis[x] = true;
            } 
        }
    }
    sort(vec.begin(), vec.end());
    memset(p, 0, sizeof(p));
    p[1][1] = p[1][2] = p[1][3] = p[1][4] = 1; 
    for(int i = 2; i < 8000; i++){
        p[i][1] = 1;
        for(int j = 1; j < 4; j++){
            if(j > i) p[i][j] = p[i][i];
            else if(j == i) p[i][j] = p[i][j - 1] + 1;
            else p[i][j] = p[i][j - 1] + p[i - j][j];
        }
    }    
} 
void solve(){
    int a = vec[n - 1], b = k;
    printf("%d\n", p[a][b]);
}

int main(){
    dowork();
    int t;
    scanf("%d", &t);
    while(t--){
        scanf("%d %d", &n, &k);
        solve();
    }
}

题目：

CZ_PROB1 - Summing to a Square Prime

#math #dynamic-programming

 

SP2={p∣p:prime∧(∃x1,x2∈Z,p=x21+x22)}SP2={p∣p:prime∧(∃x1,x2∈Z,p=x12+x22)} is the set of all primes that can be represented as the sum of two squares. The function SP2(n)SP2(n) gives the nnth prime number from the set SP2SP2. Now, given two integers nn(0<n<5010<n<501) and kk (0<k<40<k<4), find p(SP2(n),k)p(SP2(n),k) where p(a,b)p(a,b) gives the number of unordered ways to sum to the given total ‘aa’ with ‘bb’ as its largest possible part. For example: p(5,2)=3p(5,2)=3 (i.e. 2+2+12+2+1, 2+1+1+12+1+1+1, and 1+1+1+1+11+1+1+1+1). Here 55 is the total with 22 as its largest possible part.

Input

The first line gives the number of test cases TT followed by TT lines of integer pairs, nn and kk.

Constraints

• 0<T<5010<T<501
• 0<n<5010<n<501
• 1<SP2(n)<79941<SP2(n)<7994
• 0<k<40<k<4

Output

The p(SP2(n),k)p(SP2(n),k) for each nn and kk. Append a newline character to every test cases’ answer.

Example

Input:
3
2 2
3 2
5 3

Output:
3
7
85