Educational Codeforces Round 23

分析：注意两个点之间的倍数差，若为偶数则为YES，否则为NO

 1 #include "iostream"
 2 #include "cstdio"
 3 #include "cstring"
 4 #include "string"
 5 #include "cmath"
 6 using namespace std;
 7 int main()
 8 {
 9     int x1,y1,x2,y2;
10     cin>>x1>>y1>>x2>>y2;
11     int x,y;
12     cin>>x>>y;
13     int cnt1=x2-x1;
14     int cnt2=y2-y1;
15     int flag=0;
16     if(cnt1%x){
17         flag=1;
18     }
19     if(cnt2%y){
20         flag=1;
21     }
22     if(abs(abs(cnt1/x)-abs(cnt2/y))%2){
23         flag=1;
24     }
25     if(!flag){
26         cout<<"YES"<<endl;
27     }else{
28         cout<<"NO"<<endl;
29     }
30 }

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B题

分析：先看只用第一个数是否满足情况，如果不行在加入第二个数，不行在加入第三个数，如此分别统计三种情况即可

 1 #include "iostream"
 2 #include "cstdio"
 3 #include "cstring"
 4 #include "string"
 5 #include "algorithm"
 6 #include "set"
 7 #include "vector"
 8 using namespace std;
 9 const int maxn=100000+10;
10 long long a[maxn];
11 int n;
12 long long solve3(long long sum){
13     return (sum*(sum-1)*(sum-2)/6);
14 }
15 long long solve2(long long sum){
16     return (sum*(sum-1)/2);
17 }
18 int main()
19 {
20     cin>>n;
21     for(int i=0;i<n;i++)
22         cin>>a[i];
23     sort(a,a+n);
24     set<long long>h;
25     for(int i=0;i<n;i++){
26         h.insert(a[i]);
27     }
28     set<long long>::iterator it;
29     vector<long long>q;
30     for(it=h.begin();it!=h.end();it++){
31         q.push_back(*it);
32     }
33     long long cnt1=0,cnt2=0,cnt3=0;
34     for(int i=0;i<n;i++){
35         if(a[i]==q[0]){
36             cnt1++;
37         }else if(a[i]==q[1]){
38             cnt2++;
39         }else if(a[i]==q[2]){
40             cnt3++;
41         }
42     }
43     if(cnt1>=3){
44         cout<<solve3(cnt1)<<endl;
45     }else if(cnt1==2){
46         cout<<cnt2<<endl;
47     }else{
48         if(cnt2>=2){
49             cout<<solve2(cnt2)<<endl;
50         }else{
51             cout<<cnt3<<endl;
52         }
53     }
54     return 0;
55 }

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C题

分析：因为两个数的差值最大不会超过18*9=162，所以直接暴力即可

 1 #include "iostream"
 2 #include "cstdio"
 3 #include "cstring"
 4 using namespace std;
 5 long long a,b;
 6 long long solve(long long num){
 7     long long ans=0;
 8     while(num){
 9         long long mod=num%10;
10         ans+=mod;
11         num/=10;
12     }
13     return ans;
14 }
15 int main()
16 {
17     cin>>b>>a;
18     long long sum=b-a;
19     if(sum<=0){
20         cout<<"0"<<endl;
21         return 0;
22     }
23     long long cnt=0;
24     if(b-a<=162){
25         for(long long i=a;i<=b;i++){
26             long long tt=i;
27             //cout<<b-solve(tt)<<endl;
28             if((i-solve(tt))>=a)
29                 cnt++;
30         }
31         cout<<cnt<<endl;
32     }else{
33         for(long long i=a;i<=a+162;i++){
34             long long yy=i;
35             if((i-solve(yy))<a)
36                 cnt++;
37         }
38         cout<<sum-cnt+1<<endl;
39     }
40     return 0;
41 }

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