A.Plus or Minus

直接加一下

#include
#define int long long
using namespace std;int a[200005];signed main(){int n;cin>>n;while(n--){int a,b,c;cin>>a>>b>>c;if(a+b==c){cout<<"+"<

B.Grab the Candies

把奇数和偶数都加起来，再比大小

#include
#define int long long
using namespace std;int a[200005];signed main(){int t;cin>>t;while(t--){int n;cin>>n;int odd=0;int even=0;int temp=0;while(n--){cin>>temp;if(temp%2==0){even+=temp;}else{odd+=temp;}}if(even>odd){cout<<"Yes"<

C.Find and Replace

每次将字母转成数字都记录下来，然后判断和前一个是否冲突就行

#include
#define int long long
using namespace std;int a[30];
int b[30];signed main(){int t;cin>>t;while(t--){for(int i=0;i<30;i++){a[i]=0;b[i]=0;}int n;string s;cin>>n;cin>>s;int flag=1;if(n==1){cout<<"Yes"<

D.Odd Queries

在输入的时候就把他们都加起来，然后区间修改就方便多了

#include
#define int unsigned long long
using namespace std;int a[200005];signed main(){int t;cin>>t;while(t--){int n,q;cin>>n>>q;int temp;for(int i=1;i<=n;i++){cin>>temp;a[i]=a[i-1]+temp;}while(q--){int l,r,nums;cin>>l>>r>>nums;int num=a[n]-(a[r]-a[l-1])+(r-l+1)*nums;if(num%2!=0){cout<<"Yes"<

E.Interview

思路是二分查找，写的时候出了点小问题，没过，补过的；

遗憾(><)

#include
#define int long long
using namespace std;int a[200005];signed main(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int t;cin>>t;while(t--){int n;cin>>n;int temp;int ans;for(int i=1;i<=n;i++){cin>>temp;a[i]=a[i-1]+temp;}int l=1;int r=n;while(l<=r){int mid=l+(r-l)/2;cout<<"?"<<" "<>num;if(num>a[mid]-a[l-1]){ans=mid;r=mid-1;}else{l=mid+1;}}cout<<"! "<

F.Interview

写不来，好困

G1.Subsequence Addition (Easy Version)

排序，然后全加起来，判断这个数与它前面所有数的和的大小关系，若大于，则Yes,反之则No

#include
#define int unsigned long long
using namespace std;int a[5005];
int b[5005];signed main(){int t;cin>>t;while(t--){int n;cin>>n;for(int i=0;i>a[i];}sort(a,a+n);if(n==1){if(a[0]==1){cout<<"Yes"<

G2.Subsequence Addition (Hard Version)

和前一个一样只需把数组开大点就行

#include
#define int unsigned long long
using namespace std;int a[200005];
int b[200005];signed main(){int t;cin>>t;while(t--){int n;cin>>n;for(int i=0;i>a[i];}sort(a,a+n);if(n==1){if(a[0]==1){cout<<"Yes"<