2011最新个人总结ACM模板

//纪念邮票：买面值a分至b分的b-a+1张连续的邮票。

scanf("%d%d",&n,&m);s=sqrt(2*m+0.25)-0.5;i=s+1;for(k=i;k>=1;k=k-1){j=m/k-(k-1)/2;p=(int)j;if((p>0)&&(p<=n)&&((j-p)==0))printf("[%d,%d]\n",p,p+(int)k-1);}//判断是否为素数int prim(long a){int j;if(a==1)return 0;for(j=2;j<=sqrt(a);j++)if(a%j==0)break;if(j>sqrt(a))return 1;return 0 ;}//求混合积int hh(int x[],int y[],int z[]){returnx[0]*y[1]*z[2]+x[1]*y[2]*z[0]+x[2]*y[0]*z[1]-z[0]*y[1]*x[2]-x[0]*z[1] *y[2]-x[1]*y[0]*z[2];}//快速排序void quicksort(int a[],int l,int r){int i,j,t;if(l>=r)return ;i=l;j=r;t=a[i];while(i!=j){while((j>i)&&(a[j]>=t))j--;if(i<j)a[i++]=a[j];while((j>i)&&(a[i]<=t))i++;if(i<j)a[j--]=a[i];}a[i]=t;quicksort(a,l,i-1);quicksort(a,i+1,r);}//求最大公约数int gcd(int a,int b){int r;while(a&&b){if (a<b) swap(a,b);while(b>0){ r=a%b;a=b; b=r;} }return a;}//有一种数制的基数是3，权值可以取-1,0,1while(scanf("%d",&n)){if(n==0)break;memset(a,0,sizeof(a));p=0;if(n>0){while(n){ t=n%3;if(t==2){a[p]=-1;n=n/3+1;}else{a[p]=t;n=n/3;}p++;}p++;}else{ n=-n;while(n){ t=n%3;if(t==2){a[p]=-1;n=n/3+1;}else{ a[p]=t;n=n/3;}p++;}p++;for(i=0;i<=p;i++){if(a[i]!=0){a[i]=-a[i];}}}while(a[p]==0)p--;for(i=p;i>=0;i--){if(a[i]==-1)printf("-");elseprintf("%d",a[i]);}printf("\n");}//求从1到n之间有多少个素数。

（要求算法效率尽可能高）(n<=100000000) bool flag[100000000];int main(){int i,j,m,temp;scanf("%d", &temp);m = (int)sqrt(temp);for(i=2; i<=m; i++){if(flag[i]==1) continue;for(j=i*i; j<=temp; j+=i){if(flag[j]) continue;flag[j] = 1;}}for(i=2,m=0; i<=temp; i++)if(!flag[i]) m++;printf("%d\n",m);return 0;}//输入一个数n（1<=n<=10），输出n的全排列，并保证输出的最后一位为1 for(i=0;i<n;i++)a[i]=i+'1';a[i]='\0';while(1){i=n-2;while(a[i]>a[i+1] && i>=0)i--;if(i<0)break;j=n-1;while(a[j]<a[i])j--;{t=a[j];a[j]=a[i];a[i]=t;}for(j=1;j<=(n-1-i)/2;j++){ t=a[i+j];a[i+j]=a[n-j];a[n-j]=t;}if(a[n-1]=='1'){ for(i=0;i<n-1;i++){if(a[i]==':')printf("10 ");else printf("%c ",a[i]);}printf("%c\n",a[i]);//最大和子序列int p(int r){int m=0,max=0,i;for(i=1;i<=r;i++){m=m+a[i]>0?m+a[i]:0;if(m>max) max=m;}return max;}//生理周期while(p!=-1 && e!=-1 && i!=-1 && d!=-1){for(j=d+1; j<=21252; j++)if ((j-p)%23 == 0) break;for( ; j<=21252; j=j+23)if ((j-e)%28 == 0) break;for( ; j<=21252; j=j+23*28)if ((j-i)%33 == 0) break;printf("Case %d", cs);printf(": the next triple peak occurs in %d days.\n", j-d);scanf("%d%d%d%d", &p, &e, &i, &d);cs++;//邮票m种，每种有n张，用总数不超过n张的邮票，面额1开始，最多连续组成多少面额。

int a[101];int b[30000];int main(){ for(i=1;i<=m;i++)cin>>a[i];int max=0;while(1){max++;for(i=1;i<=m;i++)if((max-a[i])>=0){if(b[max]==0)b[max]=b[max-a[i]]+1;if(b[max]>b[max-a[i]]+1)b[max]=b[max-a[i]]+1;}if(b[max]==0 || b[max]>n){cout<<max-1<<endl;return 0;}}//rabbitint F[M][M];int main(){ while((m!=0) && (d!=0)){ emset(F,0,sizeof(F));for(i=0;i<m;i++)F[i][0] = i+1;for(i=m;i<=d;i++){for(j=0;j<M-1;j++){F[i][j] += F[i-1][j] + F[i-m][j];F[i][j+1] += int(F[i][j]/10);F[i][j] = F[i][j] % 10;}}flag = 0;for (i=M-1;i>=0;i--){if (F[d][i]) flag = 1;if (flag)cout<<F[d][i];}cout<<endl;cin>>m>>d;}//逆波兰式double nb(){char a[50];scanf("%s",a);if(a[0]=='+')return nb()+nb();else if(a[0]=='-') return nb()-nb();else if(a[0]=='*') return nb()*nb();else if(a[0]=='/') return nb()/nb();return atof(a); }int main(){printf("%.1f\n",nb());return 0;//放苹果const int M=101;int p[M][M];int Place(int p[][M],int m,int n){if(p[m][n]!=0) return p[m][n];if(m<n) {return p[m][n]=Place(p,m,m);}//返回的同时将该值保存下来，当后面用到时直接返回，节省时间 if(m==n){ return p[m][n]=Place(p,m,n-1)+1;}return p[m][n]=Place(p,m,n-1)+Place(p,m-n,n);}int main(){int ti,j,;for(i=0;i<M;i++)p[1][i]=p[i][1]=1;scanf("%d",&t);while(t--) { int m,n;scanf("%d %d",&m,&n);printf("%d\n",Place(p,m,n)); }return 0;}//神奇的数struct node{int p;int q;};vector<node> v;vector<node>::iterator it;int main(){int i,c,m,tmp;bool fleg;node nd;int n,k;cin>>i;while(i--){cin>>n>>k;if(n==1)cout<<k<<endl;else if(n>k)cout<<0<<endl;else{fleg=false;m=k;tmp=0;nd.p=0;nd.q=k;v.push_back(nd);do{c=(n*m+tmp)%10;tmp=(n*m+tmp)/10;m=c;nd.p=tmp;nd.q=c;for(it=v.begin();it!=v.end();it++)if(tmp==(*it).p && c==(*it).q){fleg=true;break;}if(fleg)continue;v.push_back(nd);}while(tmp!=0 || c!=k);for(it=v.end()-1;it!=v.begin()-1;it--)cout<<(*it).q;cout<<endl;v.clear();}}return 0;}//八数码问题const long maxsize = 362880;struct Node {long Number;int depth;int father;};struct queue {Node data[maxsize];int front,rear;};queue *Q[2];long fact[9] = {1,1,2,6,24,120,720,5040,40320}; //存放0至8的阶乘long mi[9] = {1,10,100,1000,10000,100000,1000000,10000000,100000000}; int exist[2][maxsize];//求结点的校验和函数int checksum(long num){int i,j,cnt[9],sum=0;char str[100];memset(cnt,0,sizeof(cnt));itoa(num,str,10);for(i=0;i<strlen(str);i++)for(j=i+1;j<strlen(str);j++)if(str[i]!='0'&&str[i]<str[j])cnt[str[j]-'0']++;for(i=1;i<=8;i++) sum+=cnt[i];return sum;}//哈希函数long hash(long m){int i,j;int f[10];long t[10];long n;n = 0;for(i=9;i>=1;i--){t[i] = m / mi[9-i] % 10;f[i] = 0;for(j=i+1;j<=9;j++) if (t[i]>t[j]) f[i]++;n += f[i]*fact[9-i];}return n;}//新结点入队void inqueue(int Q_n,long x){ if (Q[Q_n]->rear == maxsize) cout << "队列已满!";else{Q[Q_n]->rear++;Q[Q_n]->data[Q[Q_n]->rear].Number = x;Q[Q_n]->data[Q[Q_n]->rear].depth =Q[Q_n]->data[Q[Q_n]->front].depth + 1;Q[Q_n]->data[Q[Q_n]->rear].father = Q[Q_n]->front;exist[Q_n][hash(x)] = Q[Q_n]->rear;}}//头结点出队long outqueue(int Q_n){if (Q[Q_n]->front == Q[Q_n]->rear) {cout << "队列已空!";return 0;} else{Q[Q_n]->front++;return Q[Q_n]->data[Q[Q_n]->front].Number;}}void out(queue* Q,long i){if (i > 0){out(Q,Q->data[i].father);if (Q->data[i].Number > 100000000){printf("%d",Q->data[i].depth);printf(":");printf("%d\n",Q->data[i].Number);}else{printf("%d",Q->data[i].depth);printf(":0");printf("%d\n",Q->data[i].Number);}}}void anti_out(queue* Q,long i,int depth){if (i > 0){depth++;if (Q->data[i].Number > 100000000){printf("%d",depth);printf(":");printf("%d\n",Q->data[i].Number);}else{printf("%d",depth);printf(":0");printf("%d\n",Q->data[i].Number);}anti_out(Q,Q->data[i].father,depth);}}int duplicate(int Q_n,long x)//判断重复结点{//如果在相对队列中找到新扩展结点，得到答案并输出if (exist[1-Q_n][hash(x)]>=0){cout << Q[Q_n]->data[Q[Q_n]->front].depth +1 +Q[1-Q_n]->data[exist[1-Q_n][hash(x)]].depth << endl;if (Q_n == 0){out(Q[0],Q[0]->front);if (x > 100000000){printf("%d",Q[0]->data[Q[0]->front].depth+1);printf(":");printf("%d\n",x);}else{printf("%d",Q[0]->data[Q[0]->front].depth+1);printf(":0");printf("%d\n",x);}anti_out(Q[1],Q[1]->data[exist[1][hash(x)]].father,Q[0]->data [Q[0]->front].depth+1);}else{out(Q[0],exist[0][hash(x)]);anti_out(Q[1],Q[1]->front,Q[0]->data[exist[0][hash(x)]].depth );}return 2;}if (exist[Q_n][hash(x)]>=0) return 1;//在本队列中判重return 0;}int expand(int Q_n)//扩展结点{long x,new_x;int i,j,k,u;x = outqueue(Q_n);//求0的位置for(i=1;i<=9;i++) if (x / mi[9-i] % 10 == 0) {j = i; break;} for(i=1;i<=4;i++){if ((j % 3 ==0) && (i==3)) continue;if ((j % 3 ==1) && (i==2)) continue;if ((j <=3) && (i==1)) continue;if ((j >=7) && (i==4)) continue;k = j+2*i-5;u = x / mi[9-k] % 10;new_x = x - u * mi[9-k] + u * mi[9-j];switch (duplicate(Q_n,new_x)){case 0:inqueue(Q_n,new_x);break;case 2:return 0;}}return 1;}int main(){memset(exist,-1,sizeof(exist));Q[0] = (queue*)malloc(sizeof(queue));//正向初始结点入队Q1Q[0]->front = 0;Q[0]->rear = 1;cin >> Q[0]->data[1].Number;Q[0]->data[1].depth = 0;Q[0]->data[1].father = 0;exist[0][hash(Q[0]->data[1].Number)] = 1;Q[1] = (queue*)malloc(sizeof(queue));//反向初始结点入队Q2Q[1]->front = 0;Q[1]->rear = 1;cin >> Q[1]->data[1].Number;Q[1]->data[1].depth = 0;Q[1]->data[1].father = 0;exist[1][hash(Q[1]->data[1].Number)] = 1;if(checksum(Q[0]->data[1].Number)%2!=checksum(Q[1]->data[1].Number)%2) {cout << "No solution" << endl;return 0;}do{//如果Q[0]队列已空或已满if ((Q[0]->rear == Q[0]->front) || (Q[0]->rear == maxsize)){//如果Q[1]队列已空或已满if ((Q[1]->rear == Q[1]->front) || (Q[1]->rear == maxsize)) break;else if (!expand(1)) return 0;}//如果Q[1]队列已空或已满else if ((Q[1]->rear == Q[1]->front) || (Q[1]->rear == maxsize)){if (!expand(0)) return 0;}else if (Q[0]->rear <= Q[1]->rear){if (!expand(0)) return 0;}else if (!expand(1)) return 0;}while(1);return 0;}//knight骑士struct node //位置{int i;int j;};int A[9][9];int B[9][9];int d[8][2] = {1,2,1,-2,-1,2,-1,-2,2,1,2,-1,-2,1,-2,-1};vector<node> v;vector<node>::iterator it;int main(){int i,j,b,t,p=1;bool fleg;char s[3];node begin,end,tmp,tmd;while(scanf("%d",&b)==1 && b>=0){for(i=1;i<=8;i++)for(j=1;j<=8;j++){A[i][j]=0;B[i][j]=0;}for(i=0;i<b;i++){scanf("%s",&s);A[s[0]-'a'+1][s[1]-'0']=1;}scanf("%s",&s);begin.i=s[0]-'a'+1;begin.j=s[1]-'0';scanf("%s",&s);end.i=s[0]-'a'+1;end.j=s[1]-'0';v.push_back(begin);A[begin.i][begin.j]=1;B[begin.i][begin.j]=1;fleg=true;while(!v.empty() && fleg){it=v.begin();tmp=*it;v.erase(it);for(i=0;i<8;i++){tmd.i=tmp.i+d[i][0];tmd.j=tmp.j+d[i][1];if(tmd.i==end.i && tmd.j==end.j){printf("Board %d: %dmoves\n",p,B[tmp.i][tmp.j]);fleg=false;}if(tmd.i>=1 && tmd.i<=8 && tmd.j>=1 &&tmd.j<=8){if(!A[tmd.i][tmd.j]){v.push_back(tmd);A[tmd.i][tmd.j]=1;B[tmd.i][tmd.j]=B[tmp.i][tmp.j]+1;}}}}if(fleg)printf("Board %d: not reachable\n",p); v.clear();p++;}return 0;}。