这道题\(DFS\)就好了,六种情况,\(ab,ac,ba,bc,ca,cb\),我们直接枚举就可。什么?这样不会结束?用一个\(vis\)数组判断走过没有就可以了。最后排序输出即可。
\(code:\)
#include
using namespace std;
int a , b , c , tot;
int ans[100010] , vis[35][35][35];
void dfs(int x , int y , int z){
if(x == 0) ans[++tot] = z;
if(x != 0){
int xx = x , yy = y , zz = z;
y += x;
x = 0;
if(y > b) x += y - b , y = b;
if(!vis[x][y][z]){
vis[x][y][z] = 1;
dfs(x , y , z);
}
x = xx , y = yy , z = zz;
z += x;
x = 0;
if(z > c) x += z - c , z = c;
if(!vis[x][y][z]){
vis[x][y][z] = 1;
dfs(x , y , z);
}
x = xx , y = yy , z = zz;
}
if(y != 0){
int xx = x , yy = y , zz = z;
x += y;
y = 0;
if(x > a) y += x - a , x = a;
if(!vis[x][y][z]){
vis[x][y][z] = 1;
dfs(x , y , z);
}
x = xx , y = yy , z = zz;
z += y;
y = 0;
if(z > c) y += z - c , z = c;
if(!vis[x][y][z]){
vis[x][y][z] = 1;
dfs(x , y , z);
}
x = xx , y = yy , z = zz;
}
if(z != 0){
int xx = x , yy = y , zz = z;
x += z;
z = 0;
if(x > a) z += x - a , x = a;
if(!vis[x][y][z]){
vis[x][y][z] = 1;
dfs(x , y , z);
}
x = xx , y = yy , z = zz;
y += z;
z = 0;
if(y > b) z += y - b , y = b;
if(!vis[x][y][z]){
vis[x][y][z] = 1;
dfs(x , y , z);
}
x = xx , y = yy , z = zz;
}
}
int main(){
cin >> a >> b >> c;
vis[0][0][c] = 1;
dfs(0 , 0 , c);
sort(ans + 1 , ans + tot + 1);
for(int i = 1; i <= tot; i++) cout << ans[i] << " ";
return 0;
}