hdu 2131
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#include<stdio.h> #include<string.h> int main() { char ch[2],s[210]; while(scanf("%s %s",ch,s)!=EOF) { char x = ch[0]; int len = strlen(s); int cnt = 0; if( x >= 'A' && x <= 'Z' ) { for(int i = 0; i < len; ++i ) if( s[i] == x || s[i] == x+32 ) cnt++; } else { for(int i = 0; i < len; ++i ) if( s[i] == x || s[i] == x-32 ) cnt++; } printf("%.5f\n",1.0*cnt/len); } return 0; }
hdu 2132
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#include<stdio.h> int main() { int n; __int64 f[100010]={0,1}; for(__int64 i = 2; i <= 100000; ++i ) { if(i%3==0) f[i] = f[i-1]+i*i*i; else f[i] = f[i-1]+i; } while(scanf("%d",&n), n >= 0 ) { printf("%I64d\n",f[n]); } return 0; }
hdu 2133
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#include <iostream> using namespace std; int main() { int d,m,y; int month[2][13]={{0,31,28,31,30,31,30,31,31,30,31,30,31}, {0,31,29,31,30,31,30,31,31,30,31,30,31}}; while(cin>>y>>m>>d) { int sum=d; int leap=y%4==0&&y%100!=0||y%400==0; for(int i=0;i<m;i++) sum+=month[leap][i]; if(m==2) { if(leap==0&&d>28) { cout<<"illegal"<<endl; continue; } if(leap&&d>29) { cout<<"illegal"<<endl; continue; } } if((m==1||m==3||m==5||m==7||m==8||m==10||m==12)&&d>31) { cout<<"illegal"<<endl; continue; } if((m==4||m==6||m==9||m==11)&&d>30) { cout<<"illegal"<<endl; continue; } if(m==0||d==0) { cout<<"illegal"<<endl; continue; } for(int i=1;i<y;i++) { int leap=i%4==0&&i%100!=0||i%400==0; sum+=((365+leap)%7); } int k=sum%7; if(k==1) cout<<"Monday"<<endl; else if(k==2) cout<<"Tuesday"<<endl; else if(k==3) cout<<"Wednesday"<<endl; else if(k==4) cout<<"Thursday"<<endl; else if(k==5) cout<<"Friday"<<endl; else if(k==6) cout<<"Saturday"<<endl; else cout<<"Sunday"<<endl; } }
hdu 2134
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//R/r1=sqrt(1.5)，r1/r2=sqrt(2) #include<stdio.h> #include<math.h> int main() { int r; while(scanf("%d",&r), r ) { double r1 = 1.0*r/sqrt(3); double r2 = r1*sqrt(2); printf("%.3f %.3f\n",r1,r2); } return 0; }
hdu 2135
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#include<stdio.h> char a[11][11]; void change(int n,bool flag) { char b[11][11]; if(flag) { for(int i = 1; i <= n; ++i ) for(int j = 1; j <= n; ++j ) b[j][n-i+1] = a[i][j]; } else { for(int i = 1; i <= n; ++i ) for(int j = 1; j <= n; ++j ) b[n-j+1][i] = a[i][j]; } for(int i = 1; i <= n; ++i ) for(int j = 1; j <= n; ++j ) a[i][j] = b[i][j]; } int main() { int n, m; while(scanf("%d%d",&n,&m)!=EOF) { getchar(); for(int i = 1; i <= n; ++i ) scanf("%s",a[i]+1); bool flag = true; // 正 m %= 4; if( m < 0 ) flag = false, m = -m; while(m--) change(n,flag); // puts("-->") ; for(int i = 1; i <= n; ++i ) { for(int j = 1; j <= n; ++j ) printf("%c",a[i][j]); putchar('\n'); } } return 0; }
hdu 2136
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#include <stdio.h> #include <string.h> #define N 1000001 int prime[N]; int main () { for ( int i = 0; i < N; i++ ) prime[i] = -1; int num = 0; for ( int i = 2; i < N; i++ ) { if ( prime[i] == -1 ) { num++; for ( int j = i; j < N; j += i ) prime[j] = num; } } int n; while ( scanf ( "%d", &n ) == 1 ) { if ( n == 1 ) printf ( "0\n"); else printf ( "%d\n", prime[n] ); } return 0; }
hdu 2137
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题意：把一个字符串（奇数个字符）按中间字符为轴逆时针旋转n次后，输出。 mark：2WA，n居然可以为负。。。 # include <stdio.h> # include <string.h> void out0(char str[], int len){puts (str) ;} void out1(char str[], int len) { int i, j ; for (i = len-1 ; i >=0 ; i--) { for (j = 0 ; j < i ; j++) putchar (' ') ; printf ("%c\n", str[i]) ; } } void out2(char str[], int len) { int i, j ; for(i=len-1;i>=0;i--) { for (j = 0 ; j < len/2 ; j++) putchar (' ') ; printf ("%c\n", str[i]) ; } } void out3(char str[], int len) { int i, j ; for(i = len-1 ; i >= 0 ; i--) { for (j = 0 ; j < len-1-i ; j++) putchar (' ') ; printf ("%c\n", str[i]) ; } } void out4(char str[], int len) { int i ; for (i = len-1 ; i >= 0 ; i--) putchar (str[i]) ; putchar ('\n') ; } void out5(char str[], int len) { int i, j ; for (i = 0 ; i < len ; i++) { for (j = 0 ; j < len-i-1 ; j++) putchar (' ') ; printf ("%c\n", str[i]) ; } } void out6(char str[], int len) { int i, j ; for(i=0;i<len ;i++) { for (j = 0 ; j < len/2 ; j++) putchar (' ') ; printf ("%c\n", str[i]) ; } } void out7(char str[], int len) { int i, j ; for(i = 0 ; i < len ; i++) { for(j = 0 ; j < i ; j++) putchar (' ') ; printf("%c\n", str[i]) ; } } int main () { int k ; char str[100] ; while (~scanf ("%s %d%*c", str, &k)) { if (k < 0) k = (-k)%8, k = 8-k ; switch (k%8){ case 0: out0(str, strlen(str)) ; break ; case 1: out1(str, strlen(str)) ; break ; case 2: out2(str, strlen(str)) ; break ; case 3: out3(str, strlen(str)) ; break ; case 4: out4(str, strlen(str)) ; break ; case 5: out5(str, strlen(str)) ; break ; case 6: out6(str, strlen(str)) ; break ; case 7: out7(str, strlen(str)) ; break ; } } return 0 ; }
hdu 2138
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#include<stdio.h> #include<math.h> #define N 10000000 bool vis[N+10]={0}; bool isp(int n ) { int t = (int)sqrt(n); for( int i = 2; i <= t; ++i ) if(n%i==0) return false; return true; } int main() { for(int i = 2; i <= N; ++i ) { if(!vis[i]) for(int j = i+i; j <= N; j += i ) vis[j] = true; } int n,x; while(scanf("%d",&n)!=EOF) { int num = 0; for(int i = 1; i <= n; ++i ) { scanf("%d",&x ); if( x <= N ) { if( !vis[x] ) num++; } else if(isp(x)) num++; } printf("%d\n",num); } return 0; }
hdu 2139
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/* 数学题。 已知: 1^2+2^2+3^2+……+n^2 =n(n+1)(2n+1)/6 —① 那么 1^2+2^2+3^2+……+n^2+……+（2n+1）^2 =(2n+1)(n+1)(4n+3)/3 —② 又有 2^2+4^2+6^2+……+(2n)^2 =4[1^2+2^2+3^2+……+n^2]=4*① = 2n(n+1)(2n+1)/3 —③ 设所求为S 比较②和③可知 S=②-③=(2n+1)(n+1)(4n+3)/3-2n(n+1)(2n+1)/3 =(2n+1)(n+1)(2n+3)/3 —④ 因为S是2n+1项的和 把它一般化 则奇数项平方和一般公式Sn=n(n+1)(n+2)/6 */ #include<stdio.h> int main() { __int64 n; while(scanf("%I64d",&n) != EOF ) { __int64 ans = n*(n+1)*(n+2)/6; printf("%I64d\n",ans); } return 0; }