标签：think ase question limits content ext hose esc tle
1.复习代码
.复习代码
LC 322
You are given coins of different denominations and a total amount of money amount. Write a function to compute the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.
Example 1:
Input: coins =[1, 2, 5], amount =11Output:3Explanation: 11 = 5 + 5 + 1
Example 2:
Input: coins =[2], amount =3Output: -1
Note:
You may assume that you have an infinite number of each kind of coin.
//1.dp class Solution { public: int coinChange(vector<int>& coins,int amount) { //store var vector<int> dp(amount+1,INT_MAX); //initial dp[0]=0; //main func for(int coin:coins) { for(int i=coin;i<=amount;++i) { if(dp[i-coin]!=INT_MAX) dp[i]=min(dp[i],dp[i-coin]+1); } } return dp[amount]==INT_MAX?-1:dp[amount]; } }; //2.dfs+prune /*错误列表 *1.变量声明。dfs不用dp数组，且要算最小值，对变量的初始化为INT_MAX，函数体声明时都是用变量，但在调用时要直接赋值，某个变量先声明后赋值才能用在函数中 *2.void函数中的判断条件，只写for循环可能使值都无法进步。说明对终止条件的判断不足 *3.函数声明中变量元素是否有&，也是重要判断因素 <1> 变量才有可能加，0/1直接赋值的不行 <2> 函数中有加减运算的 */ class Solution { public: int coinChange(vector<int>& coins,int amount) { sort(coins.rbegin(),coins.rend()); int ans=INT_MAX; coinChange(coins,amount,0,0,ans); return ans==INT_MAX?-1:ans; } void coinChange(vector<int>& coins,int& amount,int& count,int& k,int& ans) { if(amount==0) { //计算值，全局和局部极值都应有变量保存 ans=min(ans,count); return; } if(k==coins.size()) return; const int coin=coins[k]; for(int t=amount/coin;t>=0&&count+t<ans;--t) { coinChange(coins,amount-coin*t,count+t,k+1,ans); } } };
You just bought a parcel of land that is K kilometers long; it is so narrow that, for the purposes of this problem, we will consider it to be a polyline that runs from west to east, varying in elevation. You know the elevations of the land (in meters) at K + 1 regularly spaced measurement marks M0, M1, ..., MK. These marks are 0, 1, ..., K km, respectively, from the western end.
In this region, a wooden post denotes the boundary between two adjacent parcels of land. Wooden posts can only be placed at measurement marks, and there can be at most one post at each mark. Right now, there are two posts: one at the 0 km mark, and one at the K km mark. A measurement mark with a post is considered to be part of both of the parcels it demarcates, so your parcel of land includes all measurement marks between 0 and K km, inclusive.
A parcel is considered desirable if it contains three measurement marks such that the west-most and east-most of those three marks are both strictly higher than the remaining one of the three marks, or both strictly lower than the remaining one of the three marks. People like some variation in their landscapes! Notice that these three marks are not necessarily consecutive, and the west-most and east-most of the three marks are not necessarily the west-most and east-most marks of the parcel.
Consider an example with K = 5 and M0, M1, ..., MK = 5, 6, 6, 1, 2, 4. The measurement marks with elevations 5, 2, and 4 satisfy the condition, as do the measurement marks with elevations 6, 1, and 2. However, the measurement marks with elevations 6, 6, and 1 do not satisfy the condition, nor do the measurement marks with elevations 1, 2, and 4. Any three measurement marks that satisfy the condition make the whole parcel desirable; for example, a parcel containing the measurement marks 4 7 6 7 is desirable because of the first three values.
Your parcel is desirable, but you think it may be possible to extract even more value from it! You want to add additional posts to this parcel to divide it up into multiple parcels, all of which must be desirable, since you do not want to waste any land. What is the largest number of posts you can add?
The first line of the input gives the number of test cases, T. T test cases follow. Each case begins with one line containing an integer K: the length, in kilometers, of your parcel of land. Then, there is one more line with K + 1 integers M0, M1, ..., Mk; where Mi is the elevation (in meters) at the measurement mark that is i km from the western end of your land.
For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the largest possible number of posts you can add, as described above.
Time limit: 20 seconds per test set. (10 seconds per test run.)
Memory limit: 1GB.
1 ≤ T ≤ 100.
0 ≤ Mi ≤ 1000, for all i.
(Mi - Mj) × (Mk - Mj) > 0, for some i < j < k. (Your starting parcel is desirable.)
4 ≤ K ≤ 10.
4 ≤ K ≤ 1000.
Input |
Output |
4 4 4 8 7 3 5 4 4 8 7 7 5 7 1 2 2 1 2 1 2 1 6 2 1 3 10 9 12 20 |
Case #1: 1 Case #2: 0 Case #3: 2 Case #4: 1 |
In Sample Case #1, you can add one post at 2 km to get a total of two desirable parcels. The parcel from 0 to 2 km is desirable because 4 < 8 and 8 > 7. The parcel from 2 to 4 km is desirable because 7 > 3 and 3 < 5.
In Sample Case #2, there is no way to add another post. If you added one at 1 km or 3 km, one of the parcels would include only two measurement marks and could not be desirable. If you added one at 2 km, the parcel between 0 and 2 km would be desirable, but the parcel between 2 and 4 km would not.
In Sample Case #3, posts can be added at 3 km and 5 km.
In Sample Case #4, a post can be added at 2 km. Notice that the parcel from 2 km to 6 km is desirable because 10 > 9 and 9 < 12. However, there is no way to add a second post.
/*错误列表 1.变量声明，flag反映是否需要增加后边界，应在j循环内声明，还可以减少每次初始化赋0 2.break设置 最内层k循环，若找到符合条件的及时break，相同范围的连续的不用重复计算 */ #include<bits/stdc++.h> using namespace std; int m[10001]; int main() { int T; cin>>T; for(int tmp=1;tmp<=T;++tmp) { int k; cin>>k; for(int i=0;i<=k;++i) cin>>m[i]; int ans=0,i=0,j=0; for(i=0;i<=k;i=j) for(j=i+2;j<=k;++j) { int flag=0; for(int k=i+1;k<j;++k) { if((m[k]<m[i]&&m[k]<m[j])||(m[k]>m[i]&&m[k]>m[j])) { ans++; flag=1; break; } } if(flag) break; } cout<<"Case #"<<tmp<<": "<<ans-1<<endl; } return 0; }
标签：think ase question limits content ext hose esc tle
原文地址：https://www.cnblogs.com/Marigolci/p/12306175.html
