'commit'

TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/6_2.cpp

@@ -0,0 +1,63 @@
+#include <bits/stdc++.h>
+/*
+测试用例：
+5 6
+1 0 1 0 0 0
+0 1 0 1 0 0
+1 0 0 0 1 0
+0 1 0 0 0 1
+1 0 1 0 0 0
+
+
+测试用例II：
+5 6
+1 0 1 0 0 0         2 0 4 0 0 0
+0 1 0 1 1 0         0 1 0 3 1 0
+1 0 0 0 1 0         3 0 0 0 2 0
+0 1 0 0 0 1         0 2 0 0 0 1
+1 0 1 0 0 0         1 0 1 0 0 0
+
+5 6
+1 0 1 0 0 0
+0 1 0 1 1 0
+1 0 0 0 1 0
+0 1 0 0 0 1
+1 0 1 0 0 0
+
+
+测试用例III：
+5 6
+1 0 1 0 0 0
+0 1 0 1 1 0
+1 0 0 0 1 0
+0 1 0 1 0 1
+1 0 1 0 1 0
+
+*/
+using namespace std;
+const int N = 110;
+int a[N][N];
+int res;
+
+int main() {
+    int n, m;
+    cin >> n >> m;
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= m; j++)
+            cin >> a[i][j];
+
+    //直接在原数组上递推
+    for (int i = n; i >= 1; i--)            //从倒数第二行开始
+        for (int j = 1; j <= m; j++)        //这个需要考虑一下边界问题
+            if (a[i][j])                    //只有当前位置大于0才有考虑的价值
+                a[i][j] += a[i + 1][j + 1]; //递推式
+
+    cout << "====================" << endl;
+    //打印出这个新的递推数组看看是什么情况
+    for (int i = 1; i <= n; i++) {
+        for (int j = 1; j <= m; j++)
+            cout << a[i][j] << " ";
+        cout << endl;
+    }
+    return 0;
+}