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。
时间： 2024-10-16 12:46:14