Friends number 數論 打表

題目描述：

Paula and Tai are couple. There are many stories between them. The day Paula left by airplane, Tai send one message to telephone [phone], then, everything is changing… (The story in “the snow queen”).

After a long time, Tai tells Paula, the number 220 and 284 is a couple of friends number, as they are special, all divisors of 220’s sum is 284, and all divisors of 284’s sum is 220. Can you find out there are how many couples of friends number less than 10,000. Then, how about 100,000, 200,000 and so on.

The task for you is to find out there are how many couples of friends number in given closed interval [a,b]。

Input:

There are several cases.

Each test case contains two positive integers a, b(1<= a <= b <=500,000).

Proceed to the end of file.

Output:

      For each test case, output the number of couples in the given range. The output of one test case occupied exactly one line.

原先設計了個算法。交上去2000多MS。。超時了。。後來輸出全部數據。。發現答案最多隻有28.。。打表。

這是我用於打表的代碼

/*
 * @user ipqhjjybj
 * @Time
 * @data 20130629
 */
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

#include <iostream>
#include <cmath>
#include <algorithm>
#include <numeric>
#include <utility>

#include <cstring>
#include <vector>
#include <stack>
#include <queue>
#include <map>
#include <string>
using namespace std;

#define inf 0x3f3f3f3f
#define MAXN 500010
#define clr(x,k) memset((x),(k),sizeof(x))
#define clrn(x,k) memset((x),(k),(n+1)*sizeof(int))
#define cpy(x,k) memcpy((x),(k),sizeof(x))
#define Base 10000

typedef vector<int> vi;

#define foreach(it,c) for(vi::iterator it = (c).begin();it != (c).end();++it)

#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))

int num[500020];

void getSumDivider(int n){
    int k,i,j;
    for(i = 0;i <= n;i++) num[i]=0;
    for(i = 2;i <= n;i++) num[i]+=1;
    for(i = 2;i+i <= n;i++){
        for(j = i+i,k=2;j <= n;j+=i,k++){
            num[j]+=k;
        }
    }
}
int main(){
    freopen("1008.out","w",stdout);
    int a,b,aaa,bbb,i,ans;
    getSumDivider(MAXN);
    //while(scanf("%d %d",&a,&b)!=EOF){
        a = 1;b = 500000;
        ans=0;
        for(i = a;i <= b;i++){
            aaa = num[i];
            if(aaa>b)continue;
            bbb = num[aaa];
            if(bbb == i && a <= aaa && aaa <= b && aaa != bbb)
            {
                printf("%d,%d,\n",i,aaa);
                ans++;
            }
        }
        printf("%d\n",ans/2);
    //}
    return 0;
}

這是我最終AC 的代碼：

/*
 * @user ipqhjjybj
 * @Time
 * @data 20130629
 */
#include <stdio.h>
#include <stdlib.h>
int Ans[56][2]={
    220,284,
284,220,
1184,1210,
1210,1184,
2620,2924,
2924,2620,
5020,5564,
5564,5020,
6232,6368,
6368,6232,
10744,10856,
10856,10744,
12285,14595,
14595,12285,
17296,18416,
18416,17296,
63020,76084,
66928,66992,
66992,66928,
67095,71145,
69615,87633,
71145,67095,
76084,63020,
79750,88730,
87633,69615,
88730,79750,
100485,124155,
122265,139815,
122368,123152,
123152,122368,
124155,100485,
139815,122265,
141664,153176,
142310,168730,
153176,141664,
168730,142310,
171856,176336,
176272,180848,
176336,171856,
180848,176272,
185368,203432,
196724,202444,
202444,196724,
203432,185368,
280540,365084,
308620,389924,
319550,430402,
356408,399592,
365084,280540,
389924,308620,
399592,356408,
430402,319550,
437456,455344,
455344,437456,
469028,486178,
486178,469028
};
int main(){
    int a,b,ans,i;
    while(scanf("%d %d",&a,&b)!=EOF){
        ans=0;
        for(i=0;i<56;i++)
            if(a<=Ans[i][0]&&Ans[i][0]<Ans[i][1]&&Ans[i][1]<=b)
                ans++;
        printf("%d\n",ans);
    }
    return 0;