UVa 11040 Add bricks in the wall (水题递推)

题意：给定一个金字塔，除了最后一行，每个数都等于支撑它的两个数的和，现在给奇数行的左数奇数位置，求整个金字塔。

析：很容易看出来，从下往上奇数行等于 a[i][j] = (a[i-2][j-1] - a[i][j-1] - a[i][j+1]) / 2;然后偶数行就推出来了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define print(a) printf("%d\n", (a))
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e3 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}

int a[15][15];

int main(){
    cin >> n;
    while(n--){
        memset(a, 0, sizeof(a));
        for(int i = 1; i < 10; i += 2)
            for(int j = 1; j <= i; j += 2)
                cin >> a[i][j];
        for(int i = 1; i < 10; i++)
            for(int j = 2; j <= i; j++)
                if(!a[i][j])    a[i][j] = (a[i-2][j-1] - a[i][j-1] - a[i][j+1]) / 2;
        for(int i = 2; i < 9; i += 2)
            for(int j = 1; j <= i; j++)
                a[i][j] = a[i+1][j] + a[i+1][j+1];
        for(int i = 1; i < 10; i++)
            for(int j = 1; j <= i; j++)
                if(j == i)      printf("%d\n", a[i][j]);
                else            printf("%d ", a[i][j]);
    }
    return 0;
}