RotateLeft and drop eggs programs

1-RotateLeft.cpp

@@ -0,0 +1,57 @@
+// 1-循环左移.cpp : 定义控制台应用程序的入口点。
+//
+
+#include "stdafx.h"
+#include <iostream>
+using namespace std;
+
+int N;   //the length of array
+
+void rotateleft(int *a, int n, int left) {
+	int i, j, k, change;
+
+	if(left >= n)
+		left = left % n;
+	if(left == 0)
+		return ;
+
+	for(i = 0; i + left < n; i += change) {
+		if(i + 2 * left < n)
+			change = left;
+		else
+			change = n -i - left;
+
+		for(j = i; j < i + change; ++j) {
+			a[j] = a[j] ^ a[j + left];
+			a[j + left] = a[j] ^ a[j + left];
+			a[j] = a[j] ^ a[j + left];
+		}
+
+		if(n - i == left + change) {
+			k = (left - change) % left;
+			if(k == 0)
+				return ;
+			left = k;
+		}
+	}
+}
+
+int _tmain(int argc, _TCHAR* argv[])
+{
+	int left;  //rotate left 
+	int a[] = {1, 2, 3, 4, 5, 6, 7};
+
+	cin >>left;
+
+	N = sizeof(a)/sizeof(int);
+
+	rotateleft(a, N, left);
+
+	for(int i = 0; i < N; ++i)
+		cout <<a[i] <<"   ";
+	cout <<endl;
+
+	system("pause");
+	return 0;
+}
+

2-DropEggs.cpp

@@ -0,0 +1,36 @@
+// 2-扔鸡蛋.cpp : M个鸡蛋，N层楼
+//确定在哪一层鸡蛋会破。
+//dp[m][n] = min(dp[m][n], 1 + max(dp[m - 1][i - 1], dp[m][n - i])) i = 1...n
+//典型的DP问题
+
+#include "stdafx.h"
+#include <iostream>
+using namespace std;
+
+#define N 100
+#define M 100
+
+int _tmain(int argc, _TCHAR* argv[])
+{
+	int n;
+	int dp[M + 1][N + 1];  //记录M个鸡蛋，N层楼的解
+	memset(dp, 0, sizeof(dp));
+	for(int i = 1;i <= N; ++i)   //1个鸡蛋
+		dp[1][i] = i;
+
+	for(int i = 2; i <= M; ++i){
+		for(int j = 1; j <= N; ++j) {
+			int min = 0x0FFFFFFF;
+			for(int k = 1; k <= j; ++k){
+				if(min > 1 + max(dp[i - 1][ k - 1], dp[i][ j - k]))
+					min = 1 + max(dp[i - 1][ k - 1], dp[i][ j - k]);
+			}
+			dp[i][j] = min;
+		}
+	}
+
+	cout <<dp[2][100];
+	system("pause");
+	return 0;
+}
+