UVA 1210 Sum of Consecutive Prime Numbers（数论）

标签：简单数论

UVA - 1210 Sum of Consecutive Prime Numbers Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representations does a given positive integer have? For example, the integer 53 has two representations 5 + 7 + 11 + 13 + 17 and 53 . The integer 41 has three representations 2 + 3 + 5 + 7 + 11 + 13 , 11 + 13 + 17 , and 41 . The integer 3 has only one representation, which is 3. The integer 20 has no such representations. Note that summands must be consecutive prime numbers, so neither 7 + 13 nor 3 + 5 + 5 + 7 is a valid representation for the integer 20. Your mission is to write a program that reports the number of representations for the given positive integer. Input  The input is a sequence of positive integers each in a separate line. The integers are between 2 and 10000, inclusive. The end of the input is indicated by a zero. Output  The output should be composed of lines each corresponding to an input line except the last zero. An output line includes the number of representations for the input integer as the sum of one or more consecutive prime numbers. No other characters should be inserted in the output. Sample Input  2 3 17 41 20 666 12 53 0 Sample Output  1 1 2 3 0 0 1 2 Source Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Mathematics :: Number Theory :: Prime Numbers Root :: AOAPC II: Beginning Algorithm Contests (Second Edition) (Rujia Liu) :: Chapter 10. Maths :: Exercises Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Mathematics :: Java BigInteger Class :: Bonus Features: Primality Testing Submit Status

先搞出素数表，然后预处理出素数前缀和，然后枚举区间，预处理出结果。

#include <bits/stdc++.h>
#define foreach(it,v) for(__typeof((v).begin()) it = (v).begin(); it != (v).end(); ++it)
using namespace std;
typedef long long ll;
const int maxn = 10000 + 5;
bool check[maxn];
const int SZ = 1<<20;
struct fastio{
    char inbuf[SZ];
    char outbuf[SZ];
    fastio(){
        setvbuf(stdin,inbuf,_IOFBF,SZ);
        setvbuf(stdout,outbuf,_IOFBF,SZ);
    }
}io;
vector<int> init(int n)
{
    memset(check,0,sizeof check);
    vector<int> res;
    for(int i = 2; i <= n; i++) {
        if(!check[i])res.push_back(i);
        int sz = res.size();
        for(int j = 0; j < sz; ++j) {
            if((ll)i*res[j]>n)break;
            check[i*res[j]] = true;
            if(i%res[j]==0)break;
        }
    }
    int sz = res.size();
    for(int i = 1; i < sz; i++) res[i] += res[i-1];

    vector<int> v(n+1,0);
    for(int i = 0; i < sz; i++) {
        int l = 0;
        if(i!=0) l = res[i-1];
        for(int j = i; j < sz; j++) {
            int w = res[j] - l;
            if(w>n)break;
            v[w]++;
        }
    }
    return v;
}
int main(int argc, char const *argv[])
{
    vector<int> res = init(10000);
    int n;
    while(~scanf("%d",&n)&&n) {
        printf("%d\n", res[n]);
    }
    return 0;
}

UVA 1210 Sum of Consecutive Prime Numbers（数论）

标签：简单数论

原文地址：http://blog.csdn.net/acvcla/article/details/45547427