素数判断+DFS POJ2034

题意：

构造一个n到m的数的序列，使得任意连续的2个数，3个数，4个数...d个数的和都为合数。

先打素数表，然后DFS搜索，对于每个数都判断它和前面的数的和是否满足条件。

代码：

#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include<climits>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>
using namespace std;

#define PB push_back
#define MP make_pair

#define REP(i,x,n) for(int i=x;i<(n);++i)
#define FOR(i,l,h) for(int i=(l);i<=(h);++i)
#define FORD(i,h,l) for(int i=(h);i>=(l);--i)
#define SZ(X) ((int)(X).size())
#define ALL(X) (X).begin(), (X).end()
#define RI(X) scanf("%d", &(X))
#define RII(X, Y) scanf("%d%d", &(X), &(Y))
#define RIII(X, Y, Z) scanf("%d%d%d", &(X), &(Y), &(Z))
#define DRI(X) int (X); scanf("%d", &X)
#define DRII(X, Y) int X, Y; scanf("%d%d", &X, &Y)
#define DRIII(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
#define OI(X) printf("%d",X);
#define RS(X) scanf("%s", (X))
#define MS0(X) memset((X), 0, sizeof((X)))
#define MS1(X) memset((X), -1, sizeof((X)))
#define LEN(X) strlen(X)
#define F first
#define S second
#define Swap(a, b) (a ^= b, b ^= a, a ^= b)
#define Dpoint  strcut node{int x,y}
#define cmpd int cmp(const int &a,const int &b){return a>b;}

 /*#ifdef HOME
    freopen("in.txt","r",stdin);
    #endif*/
const int MOD = 1e9+7;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<double> VD;
typedef long long LL;
typedef pair<int,int> PII;
//#define HOME

int Scan()
{
	int res = 0, ch, flag = 0;

	if((ch = getchar()) == '-')				//判断正负
		flag = 1;

	else if(ch >= '0' && ch <= '9')			//得到完整的数
		res = ch - '0';
	while((ch = getchar()) >= '0' && ch <= '9' )
		res = res * 10 + ch - '0';

	return flag ? -res : res;
}
/*----------------PLEASE-----DO-----NOT-----HACK-----ME--------------------*/

int n,m,d;
int vis[10000];
int prime[10000];
void getprime()
{
    int cnt=0;
    for(int i=2;i<=10000;i++)
        if(!vis[i])
    {
        for(int j=i*i;j<=10000;j+=i)
        vis[j]=1;

    }
}
int vis2[1005];
int ans[1005];
int ok;
void dfs(int cur,int sum)
{if(ok)
return;
if(cur==m-n+2)
{ok=1;
return;
}
for(int i=n;i<=m;i++)
if(!vis2[i])
{

    if(cur<d)
    {int ok2=1;
    int tmp=i;
    for(int j=cur-1;j>=1;j--)
        {if(!vis[ans[j]+tmp])
        {
            ok2=0;
            break;
        }
        tmp+=ans[j];

        }
    if(ok2)
    {vis2[i]=1;
        ans[cur]=i;
        dfs(cur+1,sum+i);
    vis2[i]=0;}
    }
    else
    {int ok2=1;
    int tmp=i;
    for(int j=cur-1;j>=cur+1-d;j--)
        {if(!vis[tmp+ans[j]])
    {
        ok2=0;
        break;
    }
        tmp+=ans[j];}
        if(ok2)
        {
            vis2[i]=1;
            ans[cur]=i;
            dfs(cur+1,sum-ans[cur-d+1]+i);
            vis2[i]=0;
        }
    }
    if(ok)
        return;

}

}
int main()
{getprime();
while(RIII(n,m,d)!=EOF)
{
    if(!n&&!m&&!d)
        break;
        MS0(vis2);
        ok=0;
        dfs(1,0);
    if(!ok)
    {printf("No anti-prime sequence exists.\n");
    continue;}
    REP(i,1,m-n+1)
    printf("%d,",ans[i]);
    printf("%d\n",ans[m-n+1]);

}

        return 0;
}

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