1027 Ignatius and the princess II

/* 1027 Ignatius and the princess II */<br/> # include <stdio. h> <br/> # include <string. h> <br/> # include <stdlib. h> </P> <p> int arr [9]; <br/> int Q [1, 1005]; <br/> int n, m; </P> <p> void Init () <br/>{< br/> int I; <br/> int res; <br/> for (I = 1, res = 1; I <= 7; I ++) {<br/> res * = I; <br/> arr [I] = res; <br/>}</P> <p> int find_min (int K) <br/>{< br/> int min = n + 1; <br/> int I; <br/> for (I = k + 1; I <= N; I ++) <br/>{< br/> If (Q [I] <q [Min] & Q [I]> q [k]) <br/> min = I; <br/>}< br/> return min; <br/>}</P> <p> void print () <br/>{< br/> int I; <br/> printf ("% d", Q [1]); <br/> for (I = 2; I <= N; I ++) <br/> printf ("% d", Q [I]); <br/> printf ("/N "); <br/>}</P> <p> int comp (const void * a, const void * B) <br/>{< br/> return * (int *) a-* (int *) B; <br/>}</P> <p> int comp1 (const void * a, const void * B) <br/>{< br/> return * (int *) B-* (int *) A; <br/>} </P> <p> int main () <br/> {<br/> int I, j; <br/> int min, sum; <br/> int temp; <br/> Init (); <br/> while (scanf ("% d", & N, & M) = 2) {<br/> for (I = 0; I <= n + 1; I ++) <br/> q [I] = I; <br/> for (I = 7, sum = 0; I> 0; I --) {<br/> sum + = arr [I]; <br/> If (sum <m) {<br/> min = find_min (n-I ); <br/> q [Min] ^ = Q [n-I]; <br/> q [n-I] ^ = Q [Min]; <br/> q [Min] ^ = Q [n-I]; <br/> qsort (& Q [n-I + 1], I, sizeof (INT ), comp); <br/> I ++; <br/>}< br/> Else if (sum> m) sum-= arr [I]; <br/> If (sum = m) {<br/> qsort (& Q [n-I + 1], I, sizeof (INT), comp1); <br/> Print (); <br/> break; <br/>}< br/> return 0; <br/>}</P> <p> // I am excited. It may seem an idiot to Daniel! However, you didn't know Baidu at all, and you didn't ask anyone else. You came up with a method that ranked 73 by time! <Br/> // I haven't tried this for a long time, because I have been doing new things all the time, in addition, I personally seldom think of a <br/> // method to solve the problem. Most of them are based on others' achievements ~~~~~ Next let's talk about my method ^-^ <br/> // The maximum value of question m is 10000, not very large <br/> // I thought about it, the sequence of N numbers is in full arrangement, that is, N !. <Br/> // I save 1-N to an array Q [], which is also the smallest one. <Br/> // the first small sequence of the question is to place as small as possible in front of the question, therefore, if necessary, do not change the previous number <br/> // that is, the sequence should be changed from the next 1.1 points. If M is 1, do not change the sequence, if M is 2, we need to reverse the reciprocal of two numbers <br/> // we still assume there are SEQUENCES {1, 2, 4, 5, 6} <br/> // M = 1: {1, 2, 3, 4, 5, 6} <br/> // M = 2: {1, 3, 4, 6, 5} <br/> // M = 3: {1, 2, 3, 5, 6, 4}. When M is 3, only the two reciprocal numbers cannot be changed. Then, when m is 3, it cannot be considered as 10? <Br/> // it refers to the full arrangement when we want to move the number of reciprocal I, that is, I! <M, we have to change the number of I-1. <Br/> // Well, the problem is clear now. <Br/> // now let's consider the worst case, that is, we may need to change all N numbers. Then M = n! (Note that the maximum question m is 10000 <8 !) <Br/> // if n!> M, that is, we do not need to change the number of all N, we also consider that we may want to change the number of N-1, that is, the number (n-1 )! The number of sequences is exactly the MB <br/> // But here M has two possible conditions <br/> // A). (n-1 )! = M. Then, the sequence to be searched is the maximum (least) sequence of the number of N-1 (not required in the previous number) in the reciprocal, that is to say, we need to put the n-1 numbers as big as possible before <br/> // number. This is where we sort the n-1 numbers from large to small, output the data with the preceding number. <Br/> // B). (n-1 )! <M <n !, At this time, the number of the last n also needs to change, so m, because we should try to put the number on the front, so we can find the smallest one from the number of N-1 in the last column and change the number of <br/> // nth in the last column, and then sort the number of N-1 in the last column from small to large. In this way, the number of 1st sequences is determined, and the number of N-1 records to be counted down is unknown. <br/> // here we apply the variable sum = (n-1 )! Write down the current sequence as the number of sequences, and then we consider the number of reciprocal (n-2), So repeat until when sum = m put) replace 1 with I <br/> // is actually a result of a certain number.