Codeforces Beta Round #65 (Div. 2)

话说lc小盆友前几天叫我赶紧写，现在才写完……水啊……~

 

A题……没神马好说的

#include <stdio.h> #include <string.h> char str[105]; int main() { int i,j,n,T; scanf("%d",&T); while(T--) { scanf("%s",str); n=strlen(str); if (n<=10) printf("%s/n",str); else printf("%c%d%c/n",str[0],n-2,str[n-1]); } return 0; }

B题……也没什么说的，我看数字太小了，就直接枚举，懒得用公式求……如果用公式求的话注意最后要取整的说……某人在这里wa了（偷笑……）

#include <stdio.h> int a[1005]; int main() { int i,j,n,k,t,s,l; scanf("%d%d%d",&n,&k,&t); s=n*k; for (i=0;i<n;i++) { if (i*k*1.0/s<=t*1.0/100 && (i*k*1.0+1)/s>t*1.0/100) { for (j=0;j<i;j++) { printf("%d ",k); } for (j=i;j<n;j++) { printf("%d ",0); } return 0; } } for (i=0;i<n;i++) { for (j=0;j<=k;j++) { if ((i*k+j)*1.0/s<=t*1.0/100 && ((i*k+j)*1.0+1)/s>t*1.0/100) { for (l=0;l<i;l++) { printf("%d ",k); } printf("%d ",j); for (l=i+1;l<n;l++) { printf("%d ",0); } return 0; } } } return 0; }

C题……这就是个悲剧，脑残了去求质数，然后再算n个点可能组成的k边形的种类，其实可以直接枚举k，然后再加判断n mod k==0 ，因为脑残了，就不贴代码了……

D题……很强大的模拟题……分成三种情况，1个王，2个王，没有王，如果两个王注意下输出的顺序，一个王注意是大王还是小王，两个王的时候要注意两个王不能是同一张牌…………就是堆代码了……

#include <stdio.h> #include <string.h> typedef struct { int num; int suit; }card; typedef struct { int x,y; }point; card a[20][20]; char str[20]; int vis[20][5]; point jok[2]; char suithash[4]={'C','D','H','S'}; char numhash[4]={'T','J','Q','K'}; point ret1,ret2; int n,m; int hash[15]; point res1[100]; point res2[100]; int abs(int t) { return t>0?t:-t; } char Change(int t) { char ret; if (t<=9 && t>1) ret=t+'0'; else if (t==1) ret='A'; else ret=numhash[t-10]; return ret; } int ChangeSuit(char t) { char ret; if (t=='C') ret=0; else if (t=='D') ret=1; else if (t=='H') ret=2; else ret=3; return ret; } int ChangeNum(char t) { char ret; if (t>='2' && t<='9') ret=t-'0'; else if (t=='A') ret=1; else if (t=='T') ret=10; else if (t=='J') ret=11; else if (t=='Q') ret=12; else ret=13; return ret; } bool Check1(int x,int y) { int i,j,s; s=0; memset(hash,0,sizeof(hash)); for (i=0;i<3;i++) { for (j=0;j<3;j++) { hash[a[x+i][y+j].suit]=1; } } for (i=0;i<4;i++) { s+=hash[i]; } if (s==1) return true; return false; } bool Check2(int x,int y) { int i,j,s; memset(hash,0,sizeof(hash)); for (i=0;i<3;i++) { for (j=0;j<3;j++) { hash[a[x+i][y+j].num]=1; } } s=0; for (i=0;i<14;i++) { s+=hash[i]; } if (s==9) return true; return false; } bool Check() { int i,j,up1,up2; up1=up2=0; for (i=0;i<n;i++) { for (j=0;j<m;j++) { if (i+3>n || j+3>m) continue; if (Check1(i,j)==true) { res1[up1].x=i; res1[up1++].y=j; } if (Check2(i,j)==true) { res2[up2].x=i; res2[up2++].y=j; } } } for (i=0;i<up1;i++) { for (j=i+1;j<up1;j++) { if (abs(res1[i].x-res1[j].x)>=3 || abs(res1[i].y-res1[j].y)>=3) { ret1=res1[i]; ret2=res1[j]; return true; } } } for (i=0;i<up2;i++) { for (j=i+1;j<up2;j++) { if (abs(res2[i].x-res2[j].x)>=3 || abs(res2[i].y-res2[j].y)>=3) { ret1=res2[i]; ret2=res2[j]; return true; } } } return false; } int main() { int i,j,up,t,t1,k,l; char s1,s; card tag,tag1,tagg; scanf("%d%d",&n,&m); up=0; memset(vis,0,sizeof(vis)); for (i=0;i<n;i++) { for (j=0;j<m;j++) { scanf("%s",str); if (strcmp(str,"J1")==0 || strcmp(str,"J2")==0) { jok[up].x=i; jok[up].y=j; a[i][j].num=str[1]-'0'; up++; continue; } a[i][j].suit=ChangeSuit(str[1]); a[i][j].num=ChangeNum(str[0]); vis[a[i][j].num][a[i][j].suit]=1; } } if (up==1) { tag=a[jok[0].x][jok[0].y]; for (j=1;j<=13;j++) { for (k=0;k<4;k++) { if (vis[j][k]==false) { a[jok[0].x][jok[0].y].num=j; a[jok[0].x][jok[0].y].suit=k; if (Check()==true) { printf("Solution exists./n"); t=Change(j); printf("Replace J%d with %c%c./n",tag.num,t,suithash[k]); printf("Put the first square to (%d, %d)./n",ret1.x+1,ret1.y+1); printf("Put the second square to (%d, %d)./n",ret2.x+1,ret2.y+1); return 0; } } } } } else if (up==2) { tag=a[jok[0].x][jok[0].y]; tag1=a[jok[1].x][jok[1].y]; for (i=1;i<=13;i++) { for (j=0;j<4;j++) { for (k=1;k<=13;k++) { for (l=0;l<4;l++) { if (vis[i][j]==true || vis[k][l]==true || (i==k && j==l)) continue; a[jok[0].x][jok[0].y].num=i; a[jok[0].x][jok[0].y].suit=j; a[jok[1].x][jok[1].y].num=k; a[jok[1].x][jok[1].y].suit=l; if (Check()==true) { printf("Solution exists./n"); if (tag.num==1) { t=Change(i); s=suithash[j]; t1=Change(k); s1=suithash[l]; } else { t1=Change(i); s1=suithash[j]; t=Change(k); s=suithash[l]; } printf("Replace J1 with %c%c and J2 with %c%c./n",t,s,t1,s1); printf("Put the first square to (%d, %d)./n",ret1.x+1,ret1.y+1); printf("Put the second square to (%d, %d)./n",ret2.x+1,ret2.y+1); return 0; } } } } } } else { if (Check()==true) { printf("Solution exists./n"); printf("There are no jokers./n"); printf("Put the first square to (%d, %d)./n",ret1.x+1,ret1.y+1); printf("Put the second square to (%d, %d)./n",ret2.x+1,ret2.y+1); return 0; } } printf("No solution./n"); return 0; }

E题…刚开始直接用dfs+剪枝爆，结果果断超时，减了几次未果……（以后不能偷懒…）…之后想想，写了个貌似是状态dp的玩意，把每个最后得到的原子能由哪几种提供的原子用二进制表示出来，然后用这些数进行递推，用个数组存着父状态最后打印结果的时候往回推上来就行了。代码好丑- -||

#include <stdio.h> #include <string.h> #include <algorithm> using namespace std; char begin[20][5]; char end[20][5]; char hash[101][3]={"","H","He","Li","Be","B","C","N","O","F","Ne","Na","Mg","Al","Si", "P","S","Cl","Ar","K","Ca","Sc","Ti","V","Cr","Mn","Fe","Co","Ni","Cu","Zn", "Ga","Ge","As","Se","Br","Kr","Rb","Sr","Y","Zr","Nb","Mo","Tc","Ru","Rh","Pd", "Ag","Cd","In","Sn","Sb","Te","I","Xe","Cs","Ba","La","Ce","Pr","Nd","Pm","Sm", "Eu","Gd","Tb","Dy","Ho","Er","Tm","Yb","Lu","Hf","Ta","W","Re","Os","Ir","Pt", "Au","Hg","Tl","Pb","Bi","Po","At","Rn","Fr","Ra","Ac","Th","Pa","U","Np","Pu", "Am","Cm","Bk","Cf","Es","Fm"}; int s[20]; int t[20]; int ok,n,m; int ans[20][1<<18]; int dp[18][1<<18]; int up[18]; int now[18]; int parent[18][1<<18]; int vis[18][1<<18]; int print[18][18]; int uper[18]; void Check() { int i,sum,k,p,j; for (i=0;i<(1<<n);i++) { sum=0; k=i; p=0; while(k!=0) { if (k & 1) sum+=s[p]; p++; k>>=1; } for (j=0;j<m;j++) { if (sum==t[j]) { ans[j][up[j]++]=i; } } } } void Print(int k,int p) { int let,i,j,sum,q,tag; if (k!=0) { q=dp[k-1][parent[k][p]]; sum=dp[k][p]; let=(sum & (~q)); } else let=dp[0][p]; tag=0; while(let!=0) { if (let & 1) print[k][uper[k]++]=s[tag]; tag++; let>>=1; } } int main() { int i,j,sums,sumt,p,k; scanf("%d%d",&n,&m); sums=sumt=0; for (i=0;i<n;i++) { scanf("%s",begin[i]); for (j=0;j<101;j++) { if (strcmp(hash[j],begin[i])==0) break; } s[i]=j; sums+=s[i]; } for (i=0;i<m;i++) { scanf("%s",end[i]); for (j=0;j<101;j++) { if (strcmp(hash[j],end[i])==0) break; } t[i]=j; sumt+=t[i]; } if (sums!=sumt) { printf("NO/n"); return 0; } memset(up,0,sizeof(up)); Check(); memset(vis,0,sizeof(vis)); for (i=0;i<m;i++) { if (i==0) { for (j=0;j<up[i];j++) { dp[i][j]=ans[i][j]; vis[i][ans[i][j]]=1; } now[i]=up[i]; } else { now[i]=0; for (j=0;j<now[i-1];j++) { for (k=0;k<up[i];k++) { if ((dp[i-1][j] & ans[i][k])!=0) continue; if (vis[i][dp[i-1][j] | ans[i][k]]==1) continue; dp[i][now[i]]=(dp[i-1][j] | ans[i][k]); vis[i][dp[i-1][j] | ans[i][k]]=1; parent[i][now[i]++]=j; } } } } if (now[m-1]==0) { printf("NO/n"); } else { p=0; for (i=m-1;i>=0;i--) { Print(i,p); if (i!=0) p=parent[i][p]; } printf("YES/n"); for (i=0;i<m;i++) { for (j=0;j<uper[i]-1;j++) { printf("%s+",hash[print[i][j]]); } printf("%s->%s/n",hash[print[i][j]],hash[t[i]]); } } return 0; }