辽宁省赛2010 G题 Friends Number

题目链接：http://acm.neu.edu.cn/hustoj/problem.php?id=1008

题意：数a和数b，若a的因子和（不包括a）为b，b的因子和为a，则a和b为满足条件的一对。给定一个区间，求这个区间内有多少对这样的数。

思路：打表。比赛的时候莫名其妙的没有做出来，大概是优化的时候把因子和大于MAXN时因子和赋值为-1，遍历出错。比赛的时候用了两个标记分别从前往后和从后往前遍历也是挺无聊的~

发现对二维数组，不一定要用{{a,b},{c,d}}来赋值，{a,b,c,d}的效果一样。

代码：

#include <stdio.h>

#include <math.h>

#include <string.h>

//int const MAXN = 5000000+5;

//int sum[MAXN];

//int visit[MAXN];

//void cal()

//{

//    memset(visit,0,sizeof(visit));

//    memset(sum,0,sizeof(sum));

//    int i,j;

//    int t;

//    i=1;

//    sum[1] = 0;

//    for(i=1; i<MAXN; i++){

//        j = 2*i;

//        while(j<MAXN){

//            sum[j] += i;

////            if(j==469028)

////                printf("%d %d\n",i,sum[j]);

//            j+=i;

//        }

//    }

////    printf("sum = %d\n",sum[10]);

//

////    for(i=1; i<50; i++)

////        printf("%d\n",sum[i]);

//}

//int main()

//{

//    freopen("data.txt","w",stdout);

//    cal();

////    if(sum[486178]==sum[469028])

////        printf("ok\n");

//    for(int i=1; i<MAXN; i++){

//        int t = sum[i];

//        if(t>=MAXN)

//            continue;

//        if(sum[t]==i && t!=i && visit[t]==0 && visit[i]==0){

//            printf("{%d,%d},",i,t);

//            visit[t]=1;

//            visit[i]=1;

//        }

//    }

////    printf("\n%d %d\n",sum[2803580],sum[3716164]);

//    return 0;

//}

int s[500][2] = {220,284,1184,1210,2620,2924,5020,5564,6232,6368,10744,10856,12285,14595,17296,18416,63020,76084,66928,66992,67095,71145,69615,87633,79750,88730,100485,124155,122265,139815,122368,123152,141664,153176,142310,168730,171856,176336,176272,180848,185368,203432,196724,202444,280540,365084,308620,389924,319550,430402,356408,399592,437456,455344,469028,486178,503056,514736,522405,525915,600392,669688,609928,686072,624184,691256,635624,712216,643336,652664,667964,783556,726104,796696,802725,863835,879712,901424,898216,980984,947835,1125765,998104,1043096,1077890,1099390,1154450,1189150,1156870,1292570,1175265,1438983,1185376,1286744,1280565,1340235,1328470,1483850,1358595,1486845,1392368,1464592,1466150,1747930,1468324,1749212,1511930,1598470,1669910,2062570,1798875,1870245,2082464,2090656,2236570,2429030,2652728,2941672,2723792,2874064,2728726,3077354,2739704,2928136,2802416,2947216,2803580,3716164,3276856,3721544,3606850,3892670,3786904,4300136,3805264,4006736,4238984,4314616,4246130,4488910,4259750,4445050,0};

int main()

{

    int a,b;

    int num;

    for(int i=0; i<500; i++)

        if(s[i][0]>s[i][1]){

            int temp = s[i][0];

            s[i][0] = s[i][1];

            s[i][1] = temp;

        }

 

    while(scanf("%d%d",&a,&b)!=EOF){

        int i = 0;

        int ans = 0;

        for(i=0; i<1000 ; i++){

            if(s[i][0]==0)

                break;

            if(s[i][0]>=a && s[i][1]<=b){

                ans ++;

            }

        }

        printf("%d\n",ans);

    }

    return 0;

}