Time limit 1000 ms
Memory limit 131072 kB
Paula and Tai are couple. There are many stories between them. The day Paula left by airplane, Tai send one message to telephone 2200284, 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 <=5,000,000).
Proceed to the end of file.
Each test case contains two positive integers a, b(1<= a <= b <=5,000,000).
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.
Sample Input
1 100 1 1000
Sample Output
0 1
Hint
6 is a number whose sum of all divisors is 6. 6 is not a friend number, these number is called Perfect Number.
暴力打表找出1-5000000的friends number，打表的时候记得先开方，要不然时间会耗得很长，打表得到70+组friends number
代码如下
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<algorithm> 5 using namespace std; 6 int num[][2]={ 7 220,284, 8 1184, 1210, 9 2620, 2924, 10 5020, 5564, 11 6232 ,6368, 12 10744 ,10856, 13 12285 ,14595, 14 17296 ,18416, 15 63020 ,76084, 16 66928 ,66992, 17 67095 ,71145, 18 69615 ,87633, 19 79750 ,88730, 20 100485 ,124155, 21 122265 ,139815, 22 122368 ,123152, 23 141664 ,153176, 24 142310 ,168730, 25 171856 ,176336, 26 176272 ,180848, 27 185368 ,203432, 28 196724 ,202444, 29 280540 ,365084, 30 308620 ,389924, 31 319550 ,430402, 32 356408 ,399592, 33 437456 ,455344, 34 469028 ,486178, 35 503056 ,514736, 36 522405, 525915, 37 600392, 669688, 38 609928 ,686072, 39 624184, 691256, 40 635624, 712216, 41 643336, 652664, 42 667964, 783556, 43 726104 ,796696, 44 802725 ,863835, 45 879712 ,901424, 46 898216, 980984, 47 947835, 1125765, 48 998104, 1043096, 49 1077890, 1099390, 50 1154450, 1189150, 51 1156870 ,1292570, 52 1175265 ,1438983, 53 1185376, 1286744, 54 1280565, 1340235, 55 1328470 ,1483850, 56 1358595 ,1486845, 57 1392368 ,1464592, 58 1466150,1747930, 59 1468324,1749212, 60 1511930, 1598470, 61 1669910, 2062570, 62 1798875, 1870245, 63 2082464, 2090656, 64 2236570 ,2429030, 65 2652728 ,2941672, 66 2723792 ,2874064, 67 2728726 ,3077354, 68 2739704, 2928136, 69 2802416 ,2947216, 70 2803580 ,3716164, 71 3276856 ,3721544, 72 3606850 ,3892670, 73 3786904 ,4300136, 74 3805264 ,4006736, 75 4238984 ,4314616, 76 4246130, 4488910, 77 4259750 ,4445050, 78 4482765 ,5120595, 79 4532710 ,6135962, 80 4604776 ,5162744 81 82 }; 83 84 85 86 int main() 87 { 88 int a,b; 89 int i; 90 int sum; 91 while(~scanf("%d%d",&a,&b)) 92 { 93 sum=0; 94 for(i=0;i<148;i++) 95 { 96 if(num[i][0]>=a&&num[i][1]<=b) 97 sum++; 98 if(num[i][0]>b) 99 break; 100 } 101 printf("%d ",sum); 102 } 103 return 0; 104 }