CSDN 高校俱乐部/英雄会 题目；
设数组a包含n个元素恰好是0..n - 1的一个排列，给定b[0],b[1],b[2],b[3]问有多少个0..n-1的排列a，满足(a[a[b[0]]]*b[0]+a[a[b[1]]]*b[1]+a[a[b[2]]]*b[2]+a[a[b[3]]]*b[3])%n==k ？
输入包含5个参数：N,K,B0,B1,B2,B3，其中 4<= N<12, 0 <= K,B0,B1,B2,B3 < N
单全排列超时代码：
#include <stdio.h>
#include <string.h>
int b[5];
int c[5];
int a[15];
int aa[15];
int vis[15];
int n, sum, k; int B_same_num(int b[]){
int i, j;
int p = 4;
for(i=0;i<3;++i){
if(b[i] == -1) continue;
for(j=i+1;j<4;++j)
if(b[i] == b[j]){
b[j] = -1;
--p;
}
}
return p;
} void reset(int b[]){
int i, j;
for(i=j=0; j < 4; ++j){
if(b[j] != -1)
c[i++] = b[j];
}
} void dfs(int s, int num) {
int i;
if(s == n) {
if(4 == num)
if( (c[0]*a[a[0]] + c[1]*a[a[1]] + c[2]*a[a[2]] + c[3]*a[a[3]])%n == k )
++sum;
if(3 == num)
if( (c[0]*a[0] + c[1]*a[1] + 2 * c[2]*a[2])%n == k )
++sum;
if(2 == num)
if( (c[0]*a[0] + 3 * c[1]*a[1])%n == k )
++sum;
if(1 == num)
if( (4 * c[0]*a[0])%n == k )
++sum; return;
}
for(i = 0; i < n; i++) {
if(vis[i]) continue;
vis[i] = 1;
a[s] = i;
dfs(s+1, num);
vis[i] = 0;
}
} int main(int argc, char const *argv[])
{
int t, i;
scanf("%d", &t);
while(t--){
sum = 0;
scanf("%d %d", &n, &k);
for(i=0;i<4;++i)
scanf("%d", &b[i]);
int num = B_same_num(b);
reset(b);
dfs(0, num);
printf("%d\n", sum);
}
return 0;
}
双全排列改进后AC的代码：
#include <stdio.h>
#include <string.h>
int b[5];
int c[5];
int a[15];
int d[15];
int sui[15];
int vis[15];
int vis1[15];
int arr[15];
int jiecheng[20];
int n, sum, k; void _jiecheng(){
jiecheng[0] =1;
jiecheng[1] =1;
jiecheng[2] =2;
jiecheng[3] =6;
jiecheng[4] =24;
jiecheng[5] =120;
jiecheng[6] =720;
jiecheng[7] =5040;
jiecheng[8] =40320;
jiecheng[9] =362880;
jiecheng[10] =3628800;
jiecheng[11] =39916800;
jiecheng[12] =479001600;
} void init(){
int i;
for(i=0;i<15;++i) arr[i] = 0;
} int B_same_num(int b[]){
int i, j;
int p = 4;
for(i=0;i<3;++i){
if(b[i] == -1) continue;
for(j=i+1;j<4;++j)
if(b[i] == b[j]){
b[j] = -1;
--p;
}
}
return p;
} void reset(int b[]){
int i, j;
for(i=j=0; j < 4; ++j){
if(b[j] != -1)
c[i++] = b[j];
}
} void suiji(int s, int num, int b_num) {
int i, j;
int dd;
dd = n - num - b_num; if(s == num) {
j=0;
for(i = 0; i < b_num; i++){
if(arr[a[c[i]]] == 1) continue;
else {
a[a[c[i]]] = sui[j++];
}
}
if(4 == b_num){
if( (c[0]*a[a[c[0]]] + c[1]*a[a[c[1]]] + c[2]*a[a[c[2]]] + c[3]*a[a[c[3]]])%n == k ){
sum+=jiecheng[dd];
}
}
if(3 == b_num)
if( (c[0]*a[a[c[0]]] + c[1]*a[a[c[1]]] + 2 * c[2]*a[a[c[2]]])%n == k )
sum+=jiecheng[dd];
if(2 == b_num)
if( (c[0]*a[a[c[0]]] + 3 * c[1]*a[a[c[1]]])%n == k )
sum+=jiecheng[dd];
if(1 == b_num)
if( (4 * c[0]*a[a[c[0]]])%n == k )
sum+=jiecheng[dd];
return;
}
for(i = 0; i < n; i++) {
if(vis1[i]) continue;
if(vis[i]) continue;
vis1[i] = 1;
sui[s] = i;
// printf("sui : %d\n", sui[s]);
suiji(s+1, num, b_num);
vis1[i] = 0;
}
} int arr_num(int num){
int dd = 0;
int i;
for(i=0;i<num;++i)
if(arr[a[c[i]]] == 1) continue;
else ++dd;
return dd;
} void dfs(int s, int num) {
int i, t;
if(s == num) {
init();
for(i=0;i<num;++i){
a[c[i]] = d[i];
arr[c[i]] = 1;
}
t = arr_num(num);
suiji(0, t, num);
return;
}
for(i = 0; i < n; i++) {
if(vis[i]) continue;
vis[i] = 1;
d[s] = i;
dfs(s+1, num);
vis[i] = 0;
}
} int main(int argc, char const *argv[])
{
int t, i;
scanf("%d", &t);
while(t--){
sum = 0;
_jiecheng();
scanf("%d %d", &n, &k);
for(i=0;i<4;++i)
scanf("%d", &b[i]);
int num = B_same_num(b);
reset(b);
dfs(0, num);
printf("%d\n", sum);
}
return 0;
}
精简版 代码：
#include <stdio.h>
#include <string.h> int a[12];
int b[4];
int N, K, ans;
int v[12];
int fac[14]; void factorial(){
int i;
fac[0] = 1;
for(i=1;i<=12;++i)
fac[i] = fac[i-1]*i;
} void dfs2(int s){
int t;
if(s == 4){
if ((a[a[b[0]]]*b[0]+a[a[b[1]]]*b[1]+a[a[b[2]]]*b[2]+a[a[b[3]]]*b[3])%N==K){
t = 0;
int i;
for(i=0;i<N;++i)
if(a[i] == -1) ++t;
ans += fac[t];
}
return;
}
if(a[a[b[s]]] != -1)
dfs2(s+1);
else{
int i;
for(i=0;i<N;++i)
if(!v[i]){
v[i] = 1;
a[a[b[s]]] = i;
dfs2(s+1);
v[i] = 0;
a[a[b[s]]] = -1;
}
}
} void dfs(int s){
int i;
if(s == 4){
dfs2(0);
return;
}
if(a[b[s]] != -1)
dfs(s+1);
else{
for(i=0;i<N;++i)
if(!v[i]){
v[i] = 1;
a[b[s]] = i;
dfs(s+1);
v[i] = 0;
a[b[s]] = -1;
}
}
} int main(int argc, char const *argv[])
{
int i;
scanf("%d%d%d%d%d%d", &N, &K, &b[0], &b[1], &b[2], &b[3]);
ans = 0;
memset(v, 0, sizeof(v));
for(i=0;i<12;++i) a[i] = -1;
factorial();
dfs(0);
printf("%d\n", ans);
return 0;
}
测试数据 ：
#include <stdio.h>
#include <string.h> int howmany (int N,int K,int B0,int B1,int B2,int B3)
{
if(N==4 && K==0 && B0==3 && B1==2 &&B2==1 &&B3==0) return 4;
if(N==5 && K==2 && B0==1 && B1==2 &&B2==3 &&B3==4) return 40;
if(N==6 && K==4 && B0==5 && B1==4 &&B2==3 &&B3==2) return 78;
if(N==7 && K==6 && B0==6 && B1==4 &&B2==2 &&B3==0) return 684;
if(N==8 && K==1 && B0==0 && B1==1 &&B2==2 &&B3==3) return 5454;
if(N==9 && K==3 && B0==4 && B1==8 &&B2==1 &&B3==2) return 44028;
if(N==10 && K==5 && B0==3 && B1==7 &&B2==9 &&B3==0) return 349776;
if(N==11 && K==7 && B0==3 && B1==1 &&B2==8 &&B3==9) return 3651984;
if(N==11 && K==0 && B0==0 && B1==0 &&B2==0 &&B3==0) return 39916800;
if(N==11 && K==5 && B0==4 && B1==3 &&B2==2 &&B3==1) return 3662976;
return 0;
}
PS : 和别人的代码根本无法媲美。。以后在重新改进一下这份代码！