poj1050 To the Max

真·水题。

我一开始还担心超时，自己打了个10^8的程序测时间，发现是1.06s左右，心说打个试一下，结果63msA了......

二维前缀和直接暴力枚举，O(n^4)

 1 #include <cstdio>
 2 #include <algorithm>
 3 using namespace std;
 4 
 5 int main() {
 6     int a = 0, b = 0;
 7     for(int i = 1; i <= 10010; i++) {
 8         for(int j = 1; j <= 10010; j++) {
 9             a = a + b;
10             b = b + a;
11             a -= b;
12             b -= a;
13             b = max(a, b);
14         }
15     }
16     printf("Hoing Cumor.");
17     return 0;
18 }

测试代码

 1 #include <cstdio>
 2 const int N = 103;
 3 inline void max(int &a, int b) {
 4     if(a < b) a = b;
 5     return;
 6 }
 7 int a[N][N], sum[N][N];
 8 
 9 inline void read(int &x) {
10     char c = getchar();
11     bool f = 0;
12     while(c < '0' || c > '9') {
13         if(c == '-') f = 1;
14         c = getchar();
15     }
16     while(c >= '0' && c <= '9') {
17         x = (x << 3) + (x << 1) + c - '0';
18         c = getchar();
19     }
20     if(f) x = -x;
21     return;
22 }
23 
24 int main() {
25     int n = 0;
26     read(n);
27     for(int i = 1; i <= n; i++) {
28         for(int j = 1; j <= n; j++) {
29             read(a[i][j]);
30             sum[i][j] = a[i][j] + sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1];
31         }
32     }
33     int ans = -0x7f7f7f7f;
34     for(int i = 1; i <= n; i++) {
35         for(int j = 1; j <= n; j++) {
36             for(int ii = i; ii <= n; ii++) {
37                 for(int jj = j; jj <= n; jj++) {
38                     max(ans, sum[ii][jj] - sum[i - 1][jj] - sum[ii][j - 1] + sum[i - 1][j - 1]);
39                 }
40             }
41         }
42     }
43     printf("%d", ans);
44     return 0;
45 }