紫书第三章

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例题3-1

import java.util.Scanner;import org.omg.Messaging.SyncScopeHelper;public class Main {public static void main(String[] args) {Scanner cin = new Scanner(System.in);int count=0;while (cin.hasNext()){String str;str=cin.nextLine();char[] s = str.toCharArray();for (int i=0;i<str.length();i++){ if (s[i]=='"') { if (count%2==0) System.out.printf("``"); else System.out.print("''"); count++; } else System.out.print(s[i]);}System.out.println();}}}




例题3-2

#include <bits/stdc++.h>#define gcd(a,b) __gcd(a,b)#define mset(a,x) memset(a,x,sizeof(a))#define FIN     freopen("input","r",stdin)#define FOUT    freopen("output","w",stdout)const int INF=0x3f3f3f3f;const double PI=acos(-1.0);const int MAX=1e5+10;typedef long long LL;const LL mod=1e9+7;using namespace std;char ss[100]={"`1234567890-=QWERTYUIOP[]\ASDFGHJKL;'ZXCVBNM,./"};int main (){    int n=strlen(ss);    string str;    while (getline(cin,str)){        for (int i=0;i<str.size();i++){            if (str[i]==' '){                printf (" ");continue;            }            for (int j=0;j<n;j++){                if (ss[j]==str[i]){                    printf ("%c",ss[j-1]);                    break;                }            }        }        putchar('\n');    }    return 0;}





例题3-3

#include <iostream>#include <stdlib.h>#include <stdio.h>#include <string.h>#include <math.h>#include <vector>#include <stack>#include <queue>#include <map>#include <set>#include<list>#include <bitset>#include <climits>#include <algorithm>#define gcd(a,b) __gcd(a,b)#define FIN     freopen("input","r",stdin)#define FOUT    freopen("output","w",stdout)typedef long long LL;const LL mod=1e9+7;const int INF=0x3f3f3f3f;const double PI=acos(-1.0);using namespace std;char str1[100]="A   3  HIL JM O   2TUVWXY5";char str2[100]="1SE Z  8 ";char str[1005];int main (){    //FIN;    while (scanf("%s",str)!=EOF){        int flag1=0,flag2=0;        int n=strlen(str);        for (int i=0,j=n-1;i<=j;i++,j--){            if (str[i]!=str[j]) flag1=1;            if(str[i]>='A'&&str[i]<='Z'&&(str1[str[i]-'A']!=str[j]||str1[str[i]-'A']==' ')) flag2=1;            else  if(str[i]>='1'&&str[i]<='9'&&(str2[str[i]-'1']!=str[j]||str2[str[i]-'1']==' ')) flag2=1;        }        if (flag1&&flag2) printf ("%s -- is not a palindrome.\n\n",str);        else if (!flag1&&!flag2) printf ("%s -- is a mirrored palindrome.\n\n",str);        else if (!flag1&&flag2) printf ("%s -- is a regular palindrome.\n\n",str);        else if (flag1&&!flag2) printf ("%s -- is a mirrored string.\n\n",str);    }    return 0;}




例题3-4

#include <iostream>#include <stdlib.h>#include <stdio.h>#include <string.h>#include <math.h>#include <vector>#include <stack>#include <queue>#include <map>#include <set>#include<list>#include <bitset>#include <climits>#include <algorithm>#define mset(a,x) memset(a,x,sizeof(a))#define gcd(a,b) __gcd(a,b)#define FIN     freopen("input","r",stdin)#define FOUT    freopen("output","w",stdout)typedef long long LL;const LL mod=1e9+7;const int INF=0x3f3f3f3f;const double PI=acos(-1.0);using namespace std;int a[1005];int temp[1005];int count0[20];int countt[20];int main (){    int n;    int cc=1;    //FIN;    while (scanf ("%d",&n)&&n){        mset(countt,0);        for (int i=0;i<n;i++) scanf ("%d",&a[i]),countt[a[i]]++;        printf ("Game %d:\n",cc++);        while (1){            for (int i=1;i<=9;i++) count0[i]=countt[i];            for (int i=0;i<n;i++) scanf ("%d",&temp[i]);            if (temp[0]==0) break;            int c1=0,c2=0;            for (int i=0;i<n;i++)                if (a[i]==temp[i]) c1++,count0[a[i]]--,temp[i]=0;            for (int i=0;i<n;i++)                if (count0[temp[i]]>0) c2++,count0[temp[i]]--;            printf ("    (%d,%d)\n",c1,c2);        }    }    return 0;}




例题3-5

#include <iostream>#include <stdlib.h>#include <stdio.h>#include <string.h>#include <math.h>#include <vector>#include <stack>#include <queue>#include <map>#include <set>#include<list>#include <bitset>#include <climits>#include <algorithm>#define mset(a,x) memset(a,x,sizeof(a))#define gcd(a,b) __gcd(a,b)#define FIN     freopen("input","r",stdin)#define FOUT    freopen("output","w",stdout)typedef long long LL;const LL mod=1e9+7;const int INF=0x3f3f3f3f;const double PI=acos(-1.0);using namespace std;int main (){    int t;    scanf ("%d",&t);    while(t--){        int n;        scanf ("%d",&n);        int flag=1;        for (int i=n-100;i<n;i++){            int temp=i;            int tt=i;            while (tt){                temp+=tt%10;                tt/=10;            }            if (temp==n){                printf ("%d\n",i);                flag=0;                break;            }        }        if (flag) printf ("0\n");    }    return 0;}




例题3-6

#include <stdio.h>#include <stdlib.h>#include <iostream>#include <string.h>using namespace std;int main (){int t;//freopen("data.txt","r",stdin);scanf ("%d",&t);while (t--){char str[105];scanf("%s",str);int n=strlen(str);char minn='a';char temp[105];int index;for (int i=0;i<n;i++){if (minn>str[i]){minn=str[i];index=i;}}for (int i=0;i<n;i++) temp[i]=str[(i+index)%n];for (int i=0;i<n;i++){if (str[i]==minn){int flag=0;for (int k=0;k<n;k++){if (str[(i+k)%n]<temp[k]){flag=1;break;}else if (str[(i+k)%n]>temp[k]) break;}if(flag)for (int k=0;k<n;k++) temp[k]=str[(i+k)%n];}}for (int i=0;i<n;i++) printf ("%c",temp[i]);putchar('\n');}return 0;}





习题3-1

#include <iostream>#include <stdlib.h>#include <stdio.h>#include <string.h>#include <math.h>#include <vector>#include <stack>#include <queue>#include <map>#include <set>#include<list>#include <bitset>#include <climits>#include <algorithm>#define mset(a,x) memset(a,x,sizeof(a))#define gcd(a,b) __gcd(a,b)#define FIN     freopen("input","r",stdin)#define FOUT    freopen("output","w",stdout)typedef long long LL;const LL mod=1e9+7;const int INF=0x3f3f3f3f;const double PI=acos(-1.0);using namespace std;int main (){    int t;    scanf ("%d",&t);    while (t--){        char str[100];        scanf ("%s",str);        int count0=0;        int sum=0;        for (int i=0;i<strlen(str);i++){            if (str[i]=='O') count0++,sum+=count0;            else count0=0;        }        printf ("%d\n",sum);    }    return 0;}





习题3-2

#include <iostream>#include <stdlib.h>#include <stdio.h>#include <string.h>#include <math.h>#include <vector>#include <stack>#include <queue>#include <map>#include <set>#include<list>#include <bitset>#include <climits>#include <algorithm>#define mset(a,x) memset(a,x,sizeof(a))#define gcd(a,b) __gcd(a,b)#define FIN     freopen("input","r",stdin)#define FOUT    freopen("output","w",stdout)typedef long long LL;const LL mod=1e9+7;const int INF=0x3f3f3f3f;const double PI=acos(-1.0);using namespace std;double ss[100];int main (){    ss['C']=12.01;ss['H']=1.008;ss['O']=16;ss['N']=14.01;    int t;    //FIN;    scanf ("%d",&t);    while (t--){        char str[100];        scanf ("%s",str);        int n=strlen(str);        int i=0;        double sum=0.0;        while (i<n){            if (str[i+1]!='C'&&str[i+1]!='H'&&str[i+1]!='O'&&str[i+1]!='N'&&i+1<n){                int t1=0;                double t2=ss[str[i]];                i++;                while (str[i]!='C'&&str[i]!='H'&&str[i]!='O'&&str[i]!='N'&&i<n){                    t1=t1*10+str[i]-'0';i++;                }                sum+=t1*t2;            }            else sum+=ss[str[i]],i++;        }        printf ("%.3lf\n",sum);    }    return 0;}




习题3-3

#include <iostream>#include <stdlib.h>#include <stdio.h>#include <string.h>#include <math.h>#include <vector>#include <stack>#include <queue>#include <map>#include <set>#include<list>#include <bitset>#include <climits>#include <algorithm>#define mset(a,x) memset(a,x,sizeof(a))#define gcd(a,b) __gcd(a,b)#define FIN     freopen("input","r",stdin)#define FOUT    freopen("output","w",stdout)typedef long long LL;const LL mod=1e9+7;const int INF=0x3f3f3f3f;const double PI=acos(-1.0);using namespace std;char str[100005];int count0[10];int main (){    int t;    scanf ("%d",&t);    while (t--){        int n;        scanf ("%d",&n);        mset(count0,0);        for (int i=1;i<=n;i++){            int  t=i;            while (t){                count0[t%10]++;                t/=10;            }        }        for (int i=0;i<10;i++){            if (i) putchar(' ');            printf ("%d",count0[i]);        }        putchar('\n');    }    return 0;}




习题3-4

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int main (){    int t;    scanf ("%d",&t);    int cc=0;    while (t--){    if (cc)putchar('\n');    cc++;        char str[105];        scanf ("%s",str);        int n=strlen(str);        int ff=0;        for (int i=1;i<=n/2;i++){            if (n%i==0){                int step=i;                int countx=0;                int flag=0;                while (1){                    countx++;                    int k=countx*step;                    if (k+step>n)break;                    for (int j=0;j<step;j++){                        if (str[j]!=str[j+k])  flag=1;                    }                    if (flag) break;                }                if (!flag) {                    printf ("%d\n",step);ff=1;                    break;                }            }        }        if (!ff) printf ("%d\n",n);    }    return 0;}





习题3-5

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <algorithm>using namespace std;int main (){    int cc=1;    //freopen("input","r",stdin);    while(1){        int x,y;        string mapx[10];        for (int i=0;i<5;i++){            getline(cin,mapx[i]);            if (mapx[0][0]=='Z'&&mapx[i].size()==1) goto FF;            for(int j=0;j<5;j++){                if (mapx[i][j]==' ') {                    x=i;                    y=j;                }            }        }        if(cc!=1) putchar('\n');        printf ("Puzzle #%d:\n",cc++);        int flag=0;        while (1){            char ch=getchar();            if (ch=='\n') continue;            if (ch=='0') break;            if (ch=='L'&&!flag){                int ty=y-1;                if (ty<0){                    flag=1;continue;                }                swap(mapx[x][y],mapx[x][ty]);                y=ty;            }            else if (ch=='R'&&!flag){                int ty=y+1;                if (ty>=5){                    flag=1;continue;                }                swap(mapx[x][y],mapx[x][ty]);                y=ty;            }            else if (ch=='A'&&!flag){                int tx=x-1;                if (tx<0){                    flag=1;continue;                }                swap(mapx[tx][y],mapx[x][y]);                x=tx;            }            else if(ch=='B'&&!flag){                int tx=x+1;                if (tx>=5){                    flag=1;continue;                }                swap(mapx[tx][y],mapx[x][y]);                x=tx;            }            else flag=1;        }        getchar();        if (!flag){            for (int i=0;i<5;i++){                for (int j=0;j<5;j++){                    if (j) putchar(' ');                    cout<<mapx[i][j];                }                putchar('\n');            }        }        else printf ("This puzzle has no final configuration.\n");    }    FF :;    return 0;}





习题3-6

#include <bits/stdc++.h>#define gcd(a,b) __gcd(a,b)#define mset(a,x) memset(a,x,sizeof(a))#define FIN     freopen("input","r",stdin)#define FOUT    freopen("output","w",stdout)const int INF=0x3f3f3f3f;const double PI=acos(-1.0);const int MAX=1e5+10;typedef long long LL;const LL mod=1e9+7;using namespace std;char str[50][50];int vis[50][50];int main (){    int n,m;    int cc=1;    //FIN;    while (scanf ("%d%d",&n,&m)&&n){        mset(vis,0);        if (cc!=1) putchar('\n');        printf ("puzzle #%d:\n",cc++);        for (int i=0;i<n;i++) scanf ("%s",str[i]);        printf ("Across\n");        int sum=0;        for (int i=0;i<n;i++){            for (int j=0;j<m;j++){                if (str[i][j]!='*'){                    if (i-1<0||str[i-1][j]=='*'||j-1<0||str[i][j-1]=='*'){                        int k;sum++;                        vis[i][j]=sum;                        printf ("%3d.",sum);                        for (k=j;k<m;k++){                            if (str[i][k]!='*')                                printf ("%c",str[i][k]);                            else break;                            if (i-1<0||str[i-1][k]=='*'||k-1<0||str[i][k-1]=='*') sum++,vis[i][k]=sum-1;                        }                        sum--;                        putchar('\n');                        j=k;                    }                }            }        }        printf ("Down\n");        sum=0;        for (int i=0;i<n;i++){            for (int j=0;j<m;j++){                if (str[i][j]!='*'){                    if (i-1<0||str[i-1][j]=='*'||j-1<0||str[i][j-1]=='*'){                        int k;                        printf ("%3d.",vis[i][j]);                        for (k=i;k<n;k++){                            if (str[k][j]!='*'){                                printf ("%c",str[k][j]);                                str[k][j]='*';                            }                            else break;                        }                        putchar('\n');                    }                }            }        }    }    return 0;}




习题3-7

#include <bits/stdc++.h>#define gcd(a,b) __gcd(a,b)#define mset(a,x) memset(a,x,sizeof(a))#define FIN     freopen("input","r",stdin)#define FOUT    freopen("output","w",stdout)const int INF=0x3f3f3f3f;const double PI=acos(-1.0);const int MAX=1e5+10;typedef long long LL;const LL mod=1e9+7;using namespace std;char str[55][1005];int main (){    int t;    scanf ("%d",&t);    while (t--){        int n,m;        scanf ("%d%d",&m,&n);        for (int i=0;i<m;i++) scanf ("%s",str[i]);        char temp[1005];        int sum=0;        for (int i=0;i<n;i++){            int s[10];            mset(s,0);            for (int j=0;j<m;j++){                if (str[j][i]=='A'){                    s[0]++;                }                else if (str[j][i]=='C'){                    s[1]++;                }                else if (str[j][i]=='G'){                    s[2]++;                }                else if (str[j][i]=='T'){                    s[3]++;                }            }            int maxn=0;            int index=0;            for (int k=0;k<4;k++){                if (maxn<s[k]){                    maxn=s[k];                    index=k;                }            }            if (index==0) temp[i]='A';            else if (index==1) temp[i]='C';            else if (index==2) temp[i]='G';            else if (index==3) temp[i]='T';            sum+=m-maxn;        }        for (int i=0;i<n;i++) printf("%c",temp[i]);        putchar('\n');        printf ("%d\n",sum);    }    return 0;}




习题3-8

#include <bits/stdc++.h>#define gcd(a,b) __gcd(a,b)#define mset(a,x) memset(a,x,sizeof(a))#define FIN     freopen("input","r",stdin)#define FOUT    freopen("output","w",stdout)const int INF=0x3f3f3f3f;const double PI=acos(-1.0);const int MAX=1e5+10;typedef long long LL;const LL mod=1e9+7;using namespace std;char str[55][1005];int main (){    int t;    scanf ("%d",&t);    while (t--){        int n,m;        scanf ("%d%d",&m,&n);        for (int i=0;i<m;i++) scanf ("%s",str[i]);        char temp[1005];        int sum=0;        for (int i=0;i<n;i++){            int s[10];            mset(s,0);            for (int j=0;j<m;j++){                if (str[j][i]=='A'){                    s[0]++;                }                else if (str[j][i]=='C'){                    s[1]++;                }                else if (str[j][i]=='G'){                    s[2]++;                }                else if (str[j][i]=='T'){                    s[3]++;                }            }            int maxn=0;            int index=0;            for (int k=0;k<4;k++){                if (maxn<s[k]){                    maxn=s[k];                    index=k;                }            }            if (index==0) temp[i]='A';            else if (index==1) temp[i]='C';            else if (index==2) temp[i]='G';            else if (index==3) temp[i]='T';            sum+=m-maxn;        }        for (int i=0;i<n;i++) printf("%c",temp[i]);        putchar('\n');        printf ("%d\n",sum);    }    return 0;}





习题3-9
#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;char str1[500005];char str2[500005];int main (){    while (scanf ("%s %s",str1,str2)!=EOF){        int n=strlen(str1);        int m=strlen(str2);        int k=0;        int sum=0;        for (int i=0;i<n;i++){            while (k<m){                if (str2[k]!=str1[i])k++;                else {                    sum++;k++;break;                }            }        }        if (sum==n){            printf ("Yes\n");        }        else printf ("No\n");    }    return 0;}





习题3-10

#include <bits/stdc++.h>#define gcd(a,b) __gcd(a,b)#define mset(a,x) memset(a,x,sizeof(a))#define FIN     freopen("input","r",stdin)#define FOUT    freopen("output","w",stdout)const int INF=0x3f3f3f3f;const double PI=acos(-1.0);const int MAX=1e5+10;typedef long long LL;const LL mod=1e9+7;using namespace std;struct node{    int l,r;}a[10];int cmp(node x,node y){    if (x.l==y.l) return x.r<y.r;    return x.l<y.l;}int main (){    while (1){        for (int i=0;i<6;i++){            if (scanf ("%d%d",&a[i].l,&a[i].r)==EOF) goto FF;            if (a[i].l>a[i].r) swap(a[i].l,a[i].r);        }        sort(a,a+6,cmp);        if (a[0].l!=a[1].l||a[0].r!=a[1].r||a[2].l!=a[3].l||a[2].r!=a[3].r||a[4].l!=a[5].l||a[4].r!=a[5].r){            printf ("IMPOSSIBLE\n");continue;        }        if (a[0].l==a[2].l&&((a[0].r==a[4].r&&a[2].r==a[4].l)||(a[0].r==a[4].l&&a[2].r==a[4].r)))            printf ("POSSIBLE\n");        else if (a[0].r==a[2].r&&((a[0].l==a[4].r&&a[2].l==a[4].l)||(a[0].l==a[4].l&&a[2].l==a[4].r)))            printf ("POSSIBLE\n");        else if (a[0].r==a[2].l&&((a[0].l==a[4].r&&a[2].r==a[4].l)||(a[0].l==a[4].l&&a[2].r==a[4].r)))            printf ("POSSIBLE\n");        else if(a[0].l==a[2].r&&((a[0].r==a[4].r&&a[2].l==a[4].l)||(a[0].r==a[4].l&&a[2].l==a[4].r)))            printf ("POSSIBLE\n");        else printf ("IMPOSSIBLE\n");    }    FF: ;    return 0;}





习题3-11
#include <bits/stdc++.h>#define gcd(a,b) __gcd(a,b)#define mset(a,x) memset(a,x,sizeof(a))#define FIN     freopen("input","r",stdin)#define FOUT    freopen("output","w",stdout)const int INF=0x3f3f3f3f;const double PI=acos(-1.0);const int MAX=1e5+10;typedef long long LL;const LL mod=1e9+7;using namespace std;char str1[605];char str2[605];int main (){    while(scanf ("%s%s",str1+200,str2)!=EOF){            int len1=strlen(str1+200);            int len2=strlen(str2);            for (int i=0;i<200;i++) str1[i]='1';            for (int j=len1+200;j<600;j++) str1[j]='1';            int start=200-len2;            int end=200+len1;            int minn=INF;            for (int i=start;i<end;i++){                int flag=0;                for (int j=0;j<len2;j++){                    if (str2[j]=='2'&&str1[i+j]=='2'){                        flag=1;break;                    }                }                if (!flag){                    if (len2+i<=start+len1+len2&&i<=start+len2) minn=min(minn,start+len1+len2-i);                    else if (len2+i<=start+len1+len2&&i>=start+len2&&i<=start+len1+len2) minn=min(minn,len1);                    else if (len2+i>=start+len1+len2&&i>=start+len2&&i<=start+len2+len1) minn=min(minn,len2+i-start-len2);                    else if (len2+i>=start+len1+len2&&i<=start+len2) minn=min(minn,len2);                }            }            printf ("%d\n",minn);    }    return 0;}





习题3-12

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <math.h>using namespace std;char str[1005];void cal(double &temp){int n=strlen(str);int i=0;while (str[i]!='.'){temp*=10;temp+=(str[i]-'0');i++;}i++;int countx=1;while (str[i]!='e'){temp+=pow(10,-countx)*(str[i]-'0');i++;countx++;}i++;temp=log(temp);int ttt=0;while (i<n){ttt*=10;ttt+=(str[i]-'0');i++;}temp=temp+log(10)*ttt;}int main (){//freopen("input.txt","r",stdin);while (scanf("%s",str)){if (strcmp(str,"0e0")==0) break;double temp=0;cal(temp);for (int i=0;i<=9;i++){for (int j=1;j<=30;j++){double tt=0;for (int k=0;k<i+1;k++){tt+=pow(2,-(k+1)); }double tt2=pow(2,j)-1;double temp2=log(tt)+tt2*log(2);if (fabs(temp-temp2)<1e-5){printf ("%d %d\n",i,j); goto FF;}}}FF:;}return 0;} 


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