題目大意：
稍微一看就能知道是說*和@在一排中不能相鄰，所以只需要間隔擺放就行；
我WA了三次就是有一句：
The input is terminated by a negative integer.
我沒看見，我的處理是 != -1。
AC代碼：
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <bitset>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <fstream>
#include <cstdlib>
#include <sstream>
#include <cstring>
#include <iostream>
#include <algorithm>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
//#define maxd 1010
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
#define mc(x, y) memcpy(x, y, sizeof(x))
#define ms(x,y) memset(x,y,sizeof(x))
#define rep(i,n) for(int i=0;i<(n);i++)
#define repf(i,a,b) for(int i=(a);i<=(b);i++)
#define PI pair<int,int>
//#define mapp make_pair
#define FI first
#define SE second
#define IT iterator
#define PB push_back
#define Times 10
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int > pce;
//#define N 100
const double eps = 1e-10;
const double pi = acos(-1.0);
const ll mod = 1e9+7;
const int inf = 0x3f3f3f3f;
//const ll INF = (ll)1e18+300;
const int maxd = 100000 + 10;
const int maxx = 10100;
char tower[1010][1010];
char ac[1010][1010];
int vis[1010];
int main() {
ios::sync_with_stdio(false);
int n;
while(cin >> n) {
if(n < 0) {
break;
}
ms(vis, 0);
ms(tower, 0);
ms(ac, 0);
int ans_1 = 0;
int ans_2 = 0;
for (int i = 1; i <= n; i++) {
int flag = 1;
for (int j = 1; j <= i; j++) {
cin >> tower[i][j];
if(tower[i][j] != '?') {
flag = 0;
}
if(tower[i][j] == '@') {
ans_1 ++;
}
}
if(flag) {
vis[i] = 1;
}
}
ans_2 = ans_1;
mc(ac, tower);
/*求最多，填充@多一些*/
for (int i = 1; i <= n; i++) {
if(vis[i]) {
//cout << "+++ " << i << " " << i/2 << endl;
if(i == 1) {
ans_1 ++;
continue;
}
if(i % 2 == 0){
ans_1 += i/2;
}
else {
ans_1 += i/2;
ans_1 ++;
}
}
else {
int pos;
for (int j = 1; j <= n; j++) {
if(tower[i][j] != '?') {
pos = j;
break;
}
}
for (int j = pos-1; j >= 1; j--) {
if(tower[i][j] != '?') {
continue;
}
if(tower[i][j + 1] == '*') {
tower[i][j] = '@';
ans_1 ++;
}
else {
tower[i][j] = '*';
}
}
for (int j = pos + 1; j <= i; j++) {
if(tower[i][j] != '?') {
continue;
}
if(tower[i][j - 1] == '*') {
tower[i][j] = '@';
ans_1 ++;
}
else {
tower[i][j] = '*';
}
}
}
}
/*求最少就是填充*多一些
*/
for (int i = 1; i <= n; i++) {
if(vis[i]) {
//cout << "+++ " << i << " " << i/2 << endl;
ans_2 += i/2;
}
else {
int pos;
for (int j = 1; j <= n; j++) {
if(ac[i][j] != '?') {
pos = j;
break;
}
}
for (int j = pos-1; j >= 1; j--) {
if(ac[i][j] != '?') {
continue;
}
if(ac[i][j + 1] == '*') {
ac[i][j] = '@';
ans_2 ++;
}
else {
ac[i][j] = '*';
}
}
for (int j = pos + 1; j <= i; j++) {
if(ac[i][j] != '?') {
continue;
}
if(ac[i][j - 1] == '*') {
ac[i][j] = '@';
ans_2 ++;
}
else {
ac[i][j] = '*';
}
}
}
}
// for (int i = 1; i <= n; i++) {
// for (int j = 1; j <= i; j++) {
// cout << ac[i][j] << " ";
// }cout << endl;
// }
cout << ans_1 << " " << ans_2 << endl;
}
}