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 HOMEint 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;}