UVA 10976 - Fractions Again?!

It is easy to see that for every fraction in the form (k > 0), we can always find two positive integers x and y,x ≥ y, such that:

.

Now our question is: can you write a program that counts how many such pairs of x and y there are for any givenk?

Input

Input contains no more than 100 lines, each giving a value of k (0 < k ≤ 10000).

Output

For each k, output the number of corresponding (x, y) pairs, followed by a sorted list of the values of x and y, as shown in the sample output.

Sample Input

2
12

Sample Output

2
1/2 = 1/6 + 1/3
1/2 = 1/4 + 1/4
8
1/12 = 1/156 + 1/13
1/12 = 1/84 + 1/14
1/12 = 1/60 + 1/15
1/12 = 1/48 + 1/16
1/12 = 1/36 + 1/18
1/12 = 1/30 + 1/20
1/12 = 1/28 + 1/21
1/12 = 1/24 + 1/24

更好的测试数据：

In:

Out:

2
1/3 = 1/12 + 1/4
1/3 = 1/6 + 1/6
1
1/1 = 1/2 + 1/2
3
1/4 = 1/20 + 1/5
1/4 = 1/12 + 1/6
1/4 = 1/8 + 1/8
5
1/15 = 1/240 + 1/16
1/15 = 1/90 + 1/18
1/15 = 1/60 + 1/20
1/15 = 1/40 + 1/24
1/15 = 1/30 + 1/30
8
1/92 = 1/8556 + 1/93
1/92 = 1/4324 + 1/94
1/92 = 1/2208 + 1/96
1/92 = 1/1150 + 1/100
1/92 = 1/621 + 1/108
1/92 = 1/460 + 1/115
1/92 = 1/276 + 1/138
1/92 = 1/184 + 1/184
2
1/653 = 1/427062 + 1/654
1/653 = 1/1306 + 1/1306
5
1/589 = 1/347510 + 1/590
1/589 = 1/18848 + 1/608
1/589 = 1/11780 + 1/620
1/589 = 1/1550 + 1/950
1/589 = 1/1178 + 1/1178
5
1/793 = 1/629642 + 1/794
1/793 = 1/49166 + 1/806
1/793 = 1/11102 + 1/854
1/793 = 1/4514 + 1/962
1/793 = 1/1586 + 1/1586
14
1/2384 = 1/5685840 + 1/2385
1/2384 = 1/2844112 + 1/2386
1/2384 = 1/1423248 + 1/2388
1/2384 = 1/712816 + 1/2392
1/2384 = 1/357600 + 1/2400
1/2384 = 1/179992 + 1/2416
1/2384 = 1/91188 + 1/2448
1/2384 = 1/46786 + 1/2512
1/2384 = 1/40528 + 1/2533
1/2384 = 1/24585 + 1/2640
1/2384 = 1/21456 + 1/2682
1/2384 = 1/11920 + 1/2980
1/2384 = 1/7152 + 1/3576
1/2384 = 1/4768 + 1/4768
5
1/626 = 1/392502 + 1/627
1/626 = 1/196564 + 1/628
1/626 = 1/98595 + 1/630
1/626 = 1/1878 + 1/939
1/626 = 1/1252 + 1/1252
23
1/4338 = 1/18822582 + 1/4339
1/4338 = 1/9413460 + 1/4340
1/4338 = 1/6277086 + 1/4341
1/4338 = 1/4708899 + 1/4342
1/4338 = 1/3140712 + 1/4344
1/4338 = 1/2095254 + 1/4347
1/4338 = 1/1572525 + 1/4350
1/4338 = 1/1049796 + 1/4356
1/4338 = 1/701310 + 1/4365
1/4338 = 1/527067 + 1/4374
1/4338 = 1/352824 + 1/4392
1/4338 = 1/236662 + 1/4419
1/4338 = 1/178581 + 1/4446
1/4338 = 1/120500 + 1/4500
1/4338 = 1/82422 + 1/4579
1/4338 = 1/62419 + 1/4662
1/4338 = 1/43380 + 1/4820
1/4338 = 1/30366 + 1/5061
1/4338 = 1/23859 + 1/5302
1/4338 = 1/17352 + 1/5784
1/4338 = 1/13014 + 1/6507
1/4338 = 1/10845 + 1/7230
1/4338 = 1/8676 + 1/8676
5
1/3279 = 1/10755120 + 1/3280
1/3279 = 1/3587226 + 1/3282
1/3279 = 1/1197928 + 1/3288
1/3279 = 1/13116 + 1/4372
1/3279 = 1/6558 + 1/6558
5
1/502 = 1/252506 + 1/503
1/502 = 1/126504 + 1/504
1/502 = 1/63503 + 1/506
1/502 = 1/1506 + 1/753
1/502 = 1/1004 + 1/1004
11
1/88 = 1/7832 + 1/89
1/88 = 1/3960 + 1/90
1/88 = 1/2024 + 1/92
1/88 = 1/1056 + 1/96
1/88 = 1/792 + 1/99
1/88 = 1/572 + 1/104
1/88 = 1/440 + 1/110
1/88 = 1/330 + 1/120
1/88 = 1/264 + 1/132
1/88 = 1/209 + 1/152
1/88 = 1/176 + 1/176

In:

Out:

2
1/2 = 1/6 + 1/3
1/2 = 1/4 + 1/4
8
1/12 = 1/156 + 1/13
1/12 = 1/84 + 1/14
1/12 = 1/60 + 1/15
1/12 = 1/48 + 1/16
1/12 = 1/36 + 1/18
1/12 = 1/30 + 1/20
1/12 = 1/28 + 1/21
1/12 = 1/24 + 1/24
2
1/8999 = 1/80991000 + 1/9000
1/8999 = 1/17998 + 1/17998
41
1/10000 = 1/100010000 + 1/10001
1/10000 = 1/50010000 + 1/10002
1/10000 = 1/25010000 + 1/10004
1/10000 = 1/20010000 + 1/10005
1/10000 = 1/12510000 + 1/10008
1/10000 = 1/10010000 + 1/10010
1/10000 = 1/6260000 + 1/10016
1/10000 = 1/5010000 + 1/10020
1/10000 = 1/4010000 + 1/10025
1/10000 = 1/3135000 + 1/10032
1/10000 = 1/2510000 + 1/10040
1/10000 = 1/2010000 + 1/10050
1/10000 = 1/1572500 + 1/10064
1/10000 = 1/1260000 + 1/10080
1/10000 = 1/1010000 + 1/10100
1/10000 = 1/810000 + 1/10125
1/10000 = 1/791250 + 1/10128
1/10000 = 1/635000 + 1/10160
1/10000 = 1/510000 + 1/10200
1/10000 = 1/410000 + 1/10250
1/10000 = 1/400625 + 1/10256
1/10000 = 1/322500 + 1/10320
1/10000 = 1/260000 + 1/10400
1/10000 = 1/210000 + 1/10500
1/10000 = 1/170000 + 1/10625
1/10000 = 1/166250 + 1/10640
1/10000 = 1/135000 + 1/10800
1/10000 = 1/110000 + 1/11000
1/10000 = 1/90000 + 1/11250
1/10000 = 1/88125 + 1/11280
1/10000 = 1/72500 + 1/11600
1/10000 = 1/60000 + 1/12000
1/10000 = 1/50000 + 1/12500
1/10000 = 1/42000 + 1/13125
1/10000 = 1/41250 + 1/13200
1/10000 = 1/35000 + 1/14000
1/10000 = 1/30000 + 1/15000
1/10000 = 1/26000 + 1/16250
1/10000 = 1/25625 + 1/16400
1/10000 = 1/22500 + 1/18000
1/10000 = 1/20000 + 1/20000

初次尝试的AC解法，要测试精度！应该有更好的解法，能绕过精度问题

#define RUN
#ifdef RUN#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <vector>
#include <list>
#include <cctype>
#include <algorithm>
#include <utility>
#include <math.h>using namespace std;#define LL long long
#define MAXN 10001LL xbuf[MAXN];
LL ybuf[MAXN];void printout(LL k, LL n){printf("%lld\n", n);for(int i=0; i<n; i++){printf("1/%lld = 1/%lld + 1/%lld\n", k, xbuf[i], ybuf[i]);}
}void play(LL k){LL y = 1;LL cnt = 0;memset(xbuf, 0, sizeof(xbuf));memset(ybuf, 0, sizeof(ybuf));for(y=1; y<=2*k; y++){double x = 1.0 / (1.0/k - 1.0/y);LL xInt = (LL)(x+0.5);// 经过确定精度，选用1e-4，过大或过小都不能ACif(x>0 && fabs(x-xInt)<1e-4){//printf("x:%lf, xInt:%lld, y:%lld\n", x, xInt, y);xbuf[cnt] = xInt;ybuf[cnt] = y;cnt++;}}printout(k, cnt);
}int main(){#ifndef ONLINE_JUDGEfreopen("10976.in", "r", stdin);freopen("10976.out", "w", stdout);
#endifLL k;while(scanf("%lld",&k) != EOF){play(k);}}#endif

转载于:https://www.cnblogs.com/AlgoWing/archive/2013/03/08/3189591.html