hdu 4542 小明系列故事——未知剩余系 数论

http://acm.hdu.edu.cn/showproblem.php?pid=4542

type = 0

对于整数m = p1^a1*p2^a2....pi^ai （pi为质数），其约数个数 n = (a1+1)*(a2+1)....*(ai+1)
所以对n进行分解，找到m的最小值即可。
做法是预处理出所有 m 对应的最小 n。
type = 1

也是暴力预处理打表，看代码好理解

#include<stdio.h>#include<string.h>#include<ctype.h>#include<math.h>#include<string>#include<map>#include<vector>#include<queue>#include<algorithm>using namespace std;void fre(){freopen("t.txt","r",stdin);}#define ls o<<1#define rs o<<1|1#define MS(x,y) memset(x,y,sizeof(x))#define debug(x) printf("%d",x);typedef long long LL;typedef unsigned long long UL;typedef unsigned int UI;const int MAXN = 1<<30;const int N = 50000;const LL LINF = (LL)1<<62;const double DINF = pow(2.0,62);int p[] = {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53};  //素数LL dp[18][N];//dp[i][j] 表示用p[0]~p[i]，相乘得到的“有j个约数”的最小数。int num1[N<<1],num2[N<<1];int main(){    //fre();    int T,i,j,k,cas = 0,r,type,n;    LL tem;    MS(dp,-1);    tem = 1;    for(i = 0; i < 63; ++i)    {        dp[0][i+1] = tem;        if(double(tem)*double(p[0]) > DINF) break;        tem*=2;    }    for(i = 0; i < 15; ++i)        for(j = 1; j < N; ++j)        {            if(dp[i][j]==-1) continue;            tem = p[i+1];            for(k = 1; k < 63; ++k)            {                if( double(dp[i][j])*double(tem) > DINF) break;                r = j*(k+1); if(r>=N) break;                if(dp[i+1][r] == -1 || dp[i+1][r] > dp[i][j]*tem) dp[i+1][r] = dp[i][j]*tem;                if(double(tem)*double(p[i+1]) > DINF) break;                tem*=p[i+1];            }        }    for(i = 0; i < 2*N; ++i) num1[i] = i-1,num2[i] = -1;//num1[i]表示i的非约数个数（数的范围是1～i-1），num2[i]表示“有i个约数”的最小数。    for(i = 1; i < 2*N; ++i)    {        if(num2[num1[i]]==-1) num2[num1[i]] = i;        for(j = i+i; j < 2*N; j+=i) num1[j]--;    }        scanf("%d",&T);    while(T--)    {        printf("Case %d: ",++cas);        scanf("%d%d",&type,&n);        if(type)        {            if(num2[n] == -1) printf("Illegal\n");            else printf("%d\n",num2[n]);        }        else        {            LL ans = -1;            for(i = 0; i < 16; ++i)            {                if(dp[i][n] == -1) continue;                if(ans == -1 || dp[i][n] < ans) ans = dp[i][n];            }            if(ans == -1 || ans > LINF) printf("INF\n");            else printf("%I64d\n",ans);        }    }}