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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 }
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 }
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 }
这场题出的质量相当不错，没有小结辣，滚去上课。
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原文地址：http://www.cnblogs.com/Cydiater/p/5905927.html
