BestCoder Round #77
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此处有目录↑
HDU-5650 so easy (数学)
http://acm.hdu.edu.cn/showproblem.php?pid=5650
your task is: calculate xor of all
For each test case, the first line contains a single integer number
131 2 3
0In the sample,S = {1, 2, 3}, subsets of S are: Φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}
题目大意:记f(s)为s集合中所有元素的异或,给定一个集合S,求所有f(s)的异或,s⊆S。
做的时候没有仔细分析,想了个错误的解法(n为奇数时输出所有元素的异或,否则输出0),竟然水过了,原来数据这么水...
官方题解:
设集合有n个数,则包含x的子集个数有2^(n-1)个。 那么当n > 1时,x出现了偶数次,所以其对答案的贡献就是0;当 n = 1时,其对答案的贡献是 x。
#include <cstdio>using namespace std;int T,n;long long ans,tmp;int main() { scanf("%d",&T); while(T--) { scanf("%d",&n); ans=0; for(int i=0;i<n;++i) { scanf("%I64d",&tmp); ans^=tmp; } printf("%I64d\n",(n==1)?ans:0); } return 0;}
HDU-5651 xiaoxin juju needs help (数学)
http://acm.hdu.edu.cn/showproblem.php?pid=5651
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
This summer he was working at Tencent as an intern. One day his leader came to ask xiaoxin for help. His leader gave him a string and he wanted xiaoxin to generate palindromic strings for him. Once xiaoxin generates a different palindromic string, his leader will give him a watermelon candy. The problem is how many candies xiaoxin's leader needs to buy?
For each test case, there is a single line containing a string
3aaaabba
121
题目大意:给定一行小写字母组成的字符串,求能形成多少个回文串?
若出现奇数次的字母的个数大于1,则无法组成回文串
否则只考虑左半边的情况,答案为左半边的排列数
写到一半,换了个变量名,导致初始化错误,查错半天...
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int T,len,cnt[127],tol,t,gcd,num[505],nn,deno[505],nd;long long ans;char s[1005];const long long mod=1000000007;int main() { scanf("%d",&T); while(T--) { memset(cnt,0,sizeof(cnt)); scanf("%s",s); len=strlen(s); for(int i=0;i<len;++i) ++cnt[s[i]]; tol=0; for(int i='a';i<='z';++i) if((cnt[i]&1)==1) { ++tol; --cnt[i]; } if(tol>1) { printf("0\n"); continue; } nn=len>>1; for(int i=0;i<nn;++i) num[i]=i+1; for(int i='a';i<='z';++i) {//枚举每个字母 for(int j=cnt[i]>>1;j>1;--j) { t=j; for(int k=0;k<nn&&t>1;++k) {//分子约去分母 gcd=__gcd(num[k],t); if(gcd>1) { num[k]/=gcd; t/=gcd; } } } } ans=1; for(int i=0;i<nn;++i) ans=(ans*num[i])%mod; printf("%I64d\n",ans); } return 0;}
————————————————————2题的旅游分割线————————————————————
HDU-5652 India and China Origins (二分||并查集)
http://acm.hdu.edu.cn/showproblem.php?pid=5652
Let's assume from my crude drawing that the only way to reaching from India to China or viceversa is through that grid, blue portion is the ocean and people haven't yet invented the ship. and the yellow portion is desert and has ghosts roaming around so people can't travel that way. and the black portions are the location which have mountains and white portions are plateau which are suitable for travelling. moutains are very big to get to the top, height of these mountains is infinite. So if there is mountain between two white portions you can't travel by climbing the mountain.
And at each step people can go to 4 adjacent positions.
Our archeologists have taken sample of each mountain and estimated at which point they rise up at that place. So given the times at which each mountains rised up you have to tell at which time the communication between India and China got completely cut off.
For each test case, the first line contains two space seperated integers
print -1 if these two countries still connected in the end.
Hint:
From the picture above, we can see that China and India have no communication since 4th year.
14 601101000001010000100100070 31 51 30 01 22 42 1
4
题目大意:给一个n*m的01地图(0代表能走,1代表不能走),和q个操作,第i个操作令(x,y)处为1,判断多少个操作后无法从最上面走到最下面(只能向相邻的四个方向走)。
二分:
转换成判断1是否能从最左边到达最右边时出错,判断'0'时只用判断四个方向,而判断'1'时必须判断8个方向,一直没想到,比赛时没调出来...
#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;struct Node { int x,y; Node(int xx=0,int yy=0):x(xx),y(yy) {}}u;int T,n,m,num,x[250005],y[250005];char ori[505][505],cur[505][505];bool vis[505][505];queue<Node> q;const int dx[8]={-1,-1,0,1,1,1,0,-1};const int dy[8]={0,1,1,1,0,-1,-1,-1};inline bool isInside(int xx,int yy) { return 0<=xx&&xx<n&&0<=yy&&yy<m;}bool bfs(int cc) { for(int i=0;i<n;++i) { for(int j=0;j<m;++j) { cur[i][j]=ori[i][j]; vis[i][j]=false; } } for(int i=1;i<=cc;++i) cur[x[i]][y[i]]='1'; while(!q.empty()) q.pop(); int xx,yy; for(xx=0;xx<n;++xx) {//左起的山 if(cur[xx][0]=='1') { vis[xx][0]=true; q.push(Node(xx,0)); } } while(!q.empty()) { u=q.front(); q.pop(); for(int i=0;i<8;++i) { xx=u.x+dx[i]; yy=u.y+dy[i]; if(isInside(xx,yy)&&!vis[xx][yy]&&cur[xx][yy]=='1') { if(yy==m-1)//若能到达右边则不连通 return true; vis[xx][yy]=true; q.push(Node(xx,yy)); } } } return false;}int main() { scanf("%d",&T); while(T--) { scanf("%d%d",&n,&m); for(int i=0;i<n;++i) scanf("%s",ori[i]); scanf("%d",&num); for(int i=1;i<=num;++i) scanf("%d%d",x+i,y+i); if(bfs(0)) { printf("0\n"); continue; } int l=1,r=num+1,mid; while(l<=r) { mid=(l+r)>>1; if(bfs(mid)) r=mid-1; else l=mid+1; } printf("%d\n",l>num?-1:l); } return 0;}
并查集:
#include <cstdio>#include <cstring>using namespace std;int T,n,m,num,x[250005],y[250005];int par[250005],u,xx,yy;char ori[505][505];const int sta=250002,des=250003;const int dx[4]={-1,0,1,0};const int dy[4]={0,1,0,-1};inline bool isInside(int xx,int yy) { return 0<=xx&&xx<n&&0<=yy&&yy<m;}inline int cal(int x,int y) { if(x==-1) return sta; if(x==n) return des; return x*m+y;}int getPar(int a) { if(a==par[a]) return a; return par[a]=getPar(par[a]);}inline void merg(int a,int b) { int pa=getPar(a),pb=getPar(b); if(pa!=pb) par[pa]=pb;}int main() { scanf("%d",&T); while(T--) { scanf("%d%d",&n,&m); for(int i=0;i<n;++i) scanf("%s",ori[i]); scanf("%d",&num); for(int i=1;i<=num;++i) { scanf("%d%d",x+i,y+i); ori[x[i]][y[i]]='1'; } par[sta]=sta; par[des]=des; for(int i=0;i<n;++i) {//初始化每个点与自己联通 for(int j=0;j<m;++j) { u=cal(i,j); par[u]=u; } } for(int i=0;i<n;++i) { for(int j=0;j<m;++j) { if(ori[i][j]=='0') { u=cal(i,j); for(int dir=0;dir<4;++dir) { xx=i+dx[dir]; yy=j+dy[dir]; if(xx==-1||xx==n||(isInside(xx,yy)&&ori[xx][yy]=='0')) merg(u,cal(xx,yy)); } } } } if(getPar(sta)==getPar(des)) { printf("-1\n"); continue; } while(num>0) { ori[x[num]][y[num]]='0'; u=cal(x[num],y[num]); for(int dir=0;dir<4;++dir) { xx=x[num]+dx[dir]; yy=y[num]+dy[dir]; if(xx==-1||xx==n||(isInside(xx,yy)&&ori[xx][yy]=='0')) merg(u,cal(xx,yy)); } if(getPar(sta)==getPar(des)) break; --num; } printf("%d\n",num); } return 0;}
- BestCoder Round #77
- BestCoder Round #77
- BestCoder Round #77
- BestCoder Round #77
- BestCoder Round #77
- BestCoder Round#77
- BestCoder Round #77 (div.2)
- BestCoder Round #77 (div2) B
- BestCoder Round #77 (div.2)
- BestCoder Round #77 (div2) C
- BestCoder Round #77 (div.2)
- BestCoder Round #77 (div.2) 总结
- BestCoder Round #77 (div.2)(A)
- BestCoder Round #77 (div.2)解题报告
- 【BestCoder Round #77 (div.2)】HDU5650so easy
- 【解题报告】BestCoder Round #77 (div.2)
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