bettle_king的博客

https://codeforces.com/problemset/problem/1373/C #include<bits/stdc++.h> using namespace std; typedef long long int LL; const int N=1e6+10; LL s[N]; void solve(string ss) { LL res=0,n=ss.size(); for(int i=0;i<n;i++) { if(ss[i]=='+') s[i+1].

https://codeforces.com/problemset/problem/327/A #include<bits/stdc++.h> using namespace std; int a[105],s[105],ss[105],n; int main(void) { cin>>n; for(int i=1;i<=n;i++) cin>>a[i]; for(int i=1;i<=n;i++) s[i]=s[i-1]+a[i]; int an.

https://codeforces.com/problemset/problem/363/B #include<bits/stdc++.h> using namespace std; const int N=1e5*2+10; int a[N],s[N],n,k,ans=1e9,flag; int main(void) { cin>>n>>k; for(int i=1;i<=n;i++) cin>>a[i]; for(int i=1;i<.

https://leetcode-cn.com/problems/find-the-highest-altitude/ 差分的定义，求一下前缀和就是原数组的各个数。在前缀和里求一个最大值即可。 class Solution { public: int largestAltitude(vector<int>& gain) { int ans=0,sum=0; for(int i=0;i<gain.size();i++) .

https://leetcode-cn.com/problems/sum-of-all-odd-length-subarrays/ class Solution { public: int sumOddLengthSubarrays(vector<int>& arr) { vector<int>ve; ve.push_back(0); int sum=0; for(int i=0;i<arr.size(.

https://leetcode-cn.com/problems/running-sum-of-1d-array/ class Solution { public: vector<int> runningSum(vector<int>& nums) { vector<int>ans; int sum=0; for(int i=0;i<nums.size();i++) sum+=nums[i],ans..

https://leetcode-cn.com/problems/minimum-value-to-get-positive-step-by-step-sum/ ans>=1 ans+s1>=1 -> ans>=1-s1 ans+s1+s2>=1 -> ans>=1-(s1+s2) 以此类推，找到一个最大的，可以包含所有区域的。 class Solution { public: int minStartValue(vector<int>&amp.

https://leetcode-cn.com/problems/find-pivot-index/ class Solution { public: int pivotIndex(vector<int>& nums) { int n=nums.size(); vector<int>ve; int s=0; ve.push_back(0); for(int i=0;i<nums.size();i+

https://leetcode-cn.com/problems/range-sum-query-immutable/ class NumArray { public: vector<int>s; NumArray(vector<int>& nums) { s.push_back(0); int temp=0; for(int i=0;i<nums.size();i++) { .

题目地址 只需要记录每一个数的正负号即可，求每一个前缀的正号的个数。 #include<cstdio> #include<iostream> using namespace std; typedef long long int LL; const int N=1e5*2+10; LL f[N],s[N];// f该位是正，负 s正的个数的前缀和 int main(void) { LL n; cin>>n; f[0]=1; for(int i=1;i<=n;i.

题目链接 考察的知识点： 前缀和 双指针 方法一: 纯的前缀和做法 时间复杂度O(n) 需要额外的两个前缀和数组来维护 #include<cstdio> #include<iostream> #include<algorithm> using namespace std; const int N=1e5+10; int a[N]; int b[N]; long long int s[N];//默认全选的前缀和 long long int ss[N];//按照.

https://www.acwing.com/problem/content/description/1532/ 再在后面复制相同的一份就可以了。 此时2->1 有两种方式，我们求其最小的距离。 #include<cstdio> #include<iostream> #include<algorithm> using namespace std; const int N=1e5*2+10; int a[N],s[N],n,q; int main(void) { .

https://www.acwing.com/problem/content/description/3130/ #include<cstdio> #include<iostream> using namespace std; const int N=1e6+10; int a[N],s[N],q,n; int main(void) { scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&a[i].