第一题说是只能用1块，两块，3块（我fo了），我后面差一点猜对了（苦笑）,ac 0
#include<iostream>
#include<vector>
using namespace std;
int CalulateMethodCount(int num_money);
int main() {
int num_money;
cin >> num_money;
cout << CalulateMethodCount(num_money) << endl;
return 0;
}
int CalulateMethodCount(int num_money) {
// write code here
//dp[i]表示采用1,2,...,n-1发送奖金的发放种类数
//注意：先发1再发2 先发2再发1是两种方法
/*vector<int> dp(num_money + 1, 0);
dp[0] = 1;
for (int i = 1; i <= num_money; ++i) {
for (int money = 1; money <= i; ++money) {
dp[i] += dp[i - money];
}
}
return dp.back(); */
//return pow(2,num_money-1);
//原来只能用1、2和n吗？
//n只能最后一次用
if (num_money == 1) return 1;
if (num_money == 2) return 2;
vector<int> dp(num_money + 1, 0);
dp[0] = 1; dp[1] = 1; dp[2] = 2;
for (int i = 2; i <= num_money; ++i) {
dp[i] = dp[i - 1] + dp[i - 2];
}
return dp[num_money] + 1;
} 第二题我猜测如果redo前面不是undo，则恢复了个“寂寞”，ac1
#include<iostream>
#include<vector>
#include<string>
#include<stack>
using namespace std;
int main() {
string cur;
stack<string> del;
vector<string> save;
while (cin >> cur) {
if (cur == "undo") {//撤销
if (save.empty()) continue;//无字符串可以撤销
cur = save.back();
save.pop_back();
del.push(cur);
}
else if (cur == "redo") {//恢复
if (del.empty()) continue;//无撤销的字符串可以恢复
cur = del.top();//此时del不应该为空
del.pop();
save.push_back(cur);
}
else {
save.push_back(cur);
while (not del.empty()) del.pop();//如果前一个不是撤销则恢复“空气”
}
}
for (int i = 0; i < save.size(); ++i) {
cout << save[i] << " ";
}
return 0;
} 
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