http://oj.tsinsen.com/A1109
分析:呃..不知道有没有好的数学搞法,反正直接暴力破解也就那么算了~
代码:
#include "bits/stdc++.h"
using namespace std
int p, a[9], OK, res[9]
int main() {
scanf("%d", &p)
for (int i = 0
a[i] = i + 1
do {
if (a[0] > a[5] || a[0] > a[8] || a[5] > a[8]) continue
if (a[1] > a[3] || a[6] > a[7] || a[2] > a[4]) continue
if (a[0] + a[1] + a[3] + a[5] != p) continue
if (a[5] + a[6] + a[7] + a[8] != p) continue
if (a[8] + a[4] + a[2] + a[0] != p) continue
OK += 1
if (OK <= 1) for (int i = 0
res[i] = a[i]
} while (next_permutation(a, a + 9))
if (OK <= 0) puts("NO")
else printf("%d\n%d\n%d %d\n%d %d\n%d %d %d %d\n",
OK, res[0], res[1], res[2], res[3], res[4], res[5], res[6], res[7], res[8])
return 0
}