NOIP 0924 水题记

Posted 2020-08-10

这场貌似是gcd专场？

第一题很有意思，模拟gcd的过程即可。

 1 //0924 candy
 2 //by Cydiater
 3 //2016.9.24
 4 #include <iostream>
 5 #include <cstdio>
 6 #include <cstring>
 7 #include <string>
 8 #include <algorithm>
 9 #include <queue>
10 #include <map>
11 #include <ctime>
12 #include <cmath>
13 #include <cstdlib>
14 #include <iomanip>
15 using namespace std;
16 #define ll long long
17 #define up(i,j,n)        for(int i=j;i<=n;i++)
18 #define down(i,j,n)        for(int i=j;i>=n;i--)
19 #define FILE "candy"
20 const ll MAXN=1e6+5;
21 const ll oo=10000000LL;
22 inline ll read(){
23     char ch=getchar();ll x=0,f=1;
24     while(ch>‘9‘||ch<‘0‘){if(ch==‘-‘)f=-1;ch=getchar();}
25     while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
26     return x*f;
27 }
28 ll N,M,flag,ans=0,cnt=0;
29 priority_queue<ll>A,B;
30 namespace solution{
31     ll gcd(ll a,ll b){
32         if(b==0)return a;
33         ans+=a/b*b;
34         return gcd(b,a%b);
35     }
36     void init(){
37         N=read();M=read();flag=read();
38     }
39     void slove(){
40         gcd(max(N,M),min(N,M));
41         cout<<ans<<endl;
42         if(flag==1){
43             up(i,1,M)A.push(N);
44             up(i,1,N)B.push(M);
45             while(1){
46                 if(A.size()==0||B.size()==0)break;
47                 if(A.size()>B.size())swap(A,B);
48                 ll len=A.size();
49                 up(i,1,len){
50                     int num=B.top();B.pop();
51                     int tmp=A.top();A.pop();
52                     if(tmp-num!=0)A.push(tmp-num);
53                     printf("%d ",num);
54                 }
55             }
56             puts("");
57         }
58     }
59 }
60 int main(){
61     freopen(FILE".in","r",stdin);
62     freopen(FILE".out","w",stdout);
63     using namespace solution;
64     init();
65     slove();
66     return 0;
67 }

View Code

第三题刚开始想到了$O(MN^2)$的算法，不甘心拿40分，也许70分可以用莫队xjb搞搞？

然后没想出来莫队，想出正解辣

 

其实就是玩玩前缀和什么的。

 1 //0924 gcd
 2 //by Cydiater
 3 //2016.9.24
 4 #include <iostream>
 5 #include <cstring>
 6 #include <iomanip>
 7 #include <cstdio>
 8 #include <cstdlib>
 9 #include <string>
10 #include <algorithm>
11 #include <queue>
12 #include <map>
13 #include <ctime>
14 #include <cmath>
15 using namespace std;
16 #define ll long long
17 #define up(i,j,n)        for(int i=j;i<=n;i++)
18 #define down(i,j,n)        for(int i=j;i>=n;i--)
19 #define FILE "gcd"
20 const ll MAXN=1e6+5;
21 const ll oo=0x3f3f3f3f;
22 const ll mod=1LL<<31;
23 inline ll read(){
24     char ch=getchar();ll x=0,f=1;
25     while(ch>‘9‘||ch<‘0‘){if(ch==‘-‘)f=-1;ch=getchar();}
26     while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
27     return x*f;
28 }
29 ll N,g[MAXN],c[MAXN],d[MAXN],gcd[MAXN],cd[MAXN],gc[MAXN],M,ans;
30 char s[MAXN];
31 namespace solution{
32     void init(){
33         scanf("%s",s+1);
34         N=strlen(s+1);
35     }
36     void pret(){
37         memset(g,0,sizeof(g));memset(c,0,sizeof(c));memset(d,0,sizeof(d));
38         memset(gc,0,sizeof(gc));memset(cd,0,sizeof(cd));memset(gcd,0,sizeof(gcd));
39         down(i,N,1){
40             g[i]=g[i+1]+(s[i]==‘g‘?1:0);
41             c[i]=c[i+1]+(s[i]==‘c‘?1:0);
42             d[i]=d[i+1]+(s[i]==‘d‘?1:0);
43         }
44         down(i,N,1){
45             cd[i]=(cd[i+1]+(s[i]==‘c‘?d[i]:0))%mod;
46             gc[i]=(gc[i+1]+(s[i]==‘g‘?c[i]:0))%mod;
47         }
48         down(i,N,1)gcd[i]=(gcd[i+1]+(s[i]==‘g‘?cd[i]:0))%mod;
49     }
50     void slove(){
51         M=read();
52         while(M--){
53             ll leftt=read(),rightt=read();
54             if(leftt>rightt)swap(leftt,rightt);
55             ans=(gcd[leftt]-gcd[rightt+1]+mod)%mod;
56             ll tmp_gc=(gc[leftt]-gc[rightt+1]+mod)%mod;
57             tmp_gc=(tmp_gc-((g[leftt]-g[rightt+1])*c[rightt+1])%mod+mod)%mod;
58             ans=(ans-(tmp_gc*d[rightt+1])%mod+mod)%mod;
59             ans=(ans-((g[leftt]-g[rightt+1])*cd[rightt+1])%mod+mod)%mod;
60             printf("%d\\n",(int)ans);
61         }
62     }
63 }
64 int main(){
65     freopen(FILE".in","r",stdin);
66     freopen(FILE".out","w",stdout);
67     using namespace solution;
68     init();
69     pret();
70     slove();
71     return 0;
72 }

View Code

 

T2做到现在很不爽，到现在也就照着标程抄了抄，没理解为什么。

  1 //NOIp 0924 pod
  2 //by Cydiater
  3 //2016.9.24
  4 #include <iostream>
  5 #include <cstring>
  6 #include <cstdlib>
  7 #include <cstdio>
  8 #include <queue>
  9 #include <map>
 10 #include <ctime>
 11 #include <cmath>
 12 #include <iomanip>
 13 #include <algorithm>
 14 #include <string>
 15 using namespace std;
 16 #define ll long long
 17 #define up(i,j,n)        for(int i=j;i<=n;i++)
 18 #define down(i,j,n)        for(int i=j;i>=n;i--)
 19 #define FILE "pod"
 20 const int MAXN=1e6+5;
 21 const int oo=0x3f3f3f3f;
 22 inline int read(){
 23     char ch=getchar();int x=0,f=1;
 24     while(ch>‘9‘||ch<‘0‘){if(ch==‘-‘)f=-1;ch=getchar();}
 25     while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
 26     return x*f;
 27 }
 28 int N,M,Q,LINK[MAXN],len=0,f[MAXN],f2[MAXN],tol[MAXN];
 29 struct edge{
 30     int x,y,next,v;
 31 }e[MAXN<<1];
 32 bool OK[MAXN];
 33 namespace solution{
 34     inline void insert(int x,int y,int v){e[++len].next=LINK[x];LINK[x]=len;e[len].y=y;e[len].v=v;e[len].x=x;}
 35     int gcd(int a,int b){return b==0?a:gcd(b,a%b);}
 36     int getf(int k){
 37         if(f[k]==k)        return k;
 38         f[k]=getf(f[k]);
 39         return f[k];
 40     }
 41     int getf2(int k){
 42         if(f2[k]==k)        return k;
 43         f2[k]=getf2(f2[k]);
 44         return f2[k];
 45     }
 46     void merge(int x,int y,int v){
 47         x=getf(x);y=getf(y);
 48         if(tol[x]!=0)v=gcd(v,tol[x]);
 49         if(tol[y]!=0)v=gcd(v,tol[y]);
 50         tol[x]=v;f[y]=x;
 51     }
 52     int merge2(int x,int y,int v){
 53         if(v==1){
 54             if(getf2(x)==getf2(y))        return 0;
 55             f2[getf2(x)]=getf2(y+N);
 56             f2[getf2(y)]=getf2(x+N);
 57         }else{
 58             if(getf2(x)==getf2(y+N))    return 0;
 59             f2[getf2(x)]=getf2(y);
 60             f2[getf2(x+N)]=getf2(y+N);
 61         }
 62         return 1;
 63     }
 64     void init(){
 65         N=read();M=read();Q=read();
 66         memset(tol,0,sizeof(tol));
 67         up(i,1,N){
 68             f[i]=i;f2[i]=i;
 69             f2[i+N]=i+N;
 70         }
 71         up(i,1,M){
 72             int x=read(),y=read(),v=read();
 73             insert(x,y,v);insert(y,x,v);
 74             merge(x,y,v);
 75         }
 76         up(i,1,M<<1){
 77             int x=e[i].x,y=e[i].y,v=e[i].v;
 78             int d=tol[getf(e[i].x)];
 79             int flag=merge2(x,y,(v/d)%2);
 80             if(!flag)OK[getf(x)]=1;
 81             i+=1;
 82         }
 83     }
 84     void slove(){
 85         while(Q--){
 86             int x=read(),y=read(),K=read();
 87             if(getf(x)!=getf(y))        puts("NIE");//unconnect
 88             else{
 89                 int d=gcd(K,tol[getf(x)]);
 90                 if((K/d)&1)                            puts("0");
 91                 else{
 92                     if(OK[getf(x)])                    puts("0");
 93                     else{
 94                         if(getf2(x)==getf2(y))        puts("0");//same color
 95                         else                          printf("%d\\n",d);
 96                     }
 97                 }
 98             }
 99         }
100     }
101 }
102 int main(){
103     freopen(FILE".in","r",stdin);
104     freopen(FILE".out","w",stdout);
105     using namespace solution;
106     init();
107     slove();
108     return 0;
109 }

View Code

 

这场题出的质量相当不错，没有小结辣，滚去上课。