• 杭电oj2064、2067、2068、2073、2076-2078、2080、2083-2085


    2064  汉诺塔III

     1 #include<stdio.h>
     2 
     3 int main(){
     4     int n,i;
     5     _int64 s[40];
     6     while(~scanf("%d",&n)){
     7         s[1] = 2;
     8         for(i=2;i<=n;i++){
     9             s[i] = 3*s[i-1] + 2;
    10         }
    11         printf("%I64d
    ",s[n]);
    12     }
    13 
    14 }

    2067  小兔的棋盘

     1 #include<stdio.h>
     2 
     3 
     4 int main(){
     5     int n,i,k,j=0;
     6     _int64 s[40] = {0};
     7     s[1] = 1;s[0] = 1;
     8     for(i=2;i<=35;i++){
     9         for(k=0;k<i;k++){
    10             s[i] += s[k]*s[i-1-k];
    11         }
    12     }
    13     while(~scanf("%d",&n)){
    14         if(n==-1){break;}
    15         
    16         printf("%d %d %I64d
    ",++j,n,s[n]*2);
    17     }
    18 
    19 }

    卡特兰数的推导公式:

    2068  RPG的错排

     1 #include<stdio.h>
     2 _int64 c(int a,int b){
     3     _int64 mul1=1,mul2=1,i;
     4     for(i=a;i>0;i--){
     5         mul1 *= i;
     6         mul2 *= b;
     7         b--;
     8     }
     9     return mul2/mul1;
    10 }
    11 
    12 int main(){
    13     int n,i;
    14     _int64 s[15],sum;
    15     s[1] = 0;s[2] = 1;
    16     for(i=3;i<15;i++){
    17         s[i] = (i-1)*(s[i-1]+s[i-2]);
    18     }
    19     while(~scanf("%d",&n)){
    20         if(n==0){break;}
    21         sum = 1;
    22         for(i=1;i<=n/2;i++){
    23             sum += c(i,n)*s[i];
    24         }
    25         printf("%I64d
    ",sum);
    26         
    27     }
    28 
    29 }

    2073  无限的路

     1 #include<stdio.h>
     2 #include<math.h>
     3 double len(int x,int y){
     4     int i,j;
     5     double len = 0,l;
     6     double t = sqrt(2);
     7     for(i=1;i<x+y;i++){
     8         len = len + t*i; 
     9     }
    10     for(i=0;i<x+y;i++){
    11         len = len + sqrt(i*i+(i+1)*(i+1));
    12     }
    13     len = len + t*x;
    14     return len;
    15 }
    16 
    17 int main(){
    18     int n,x1,y1,x2,y2;
    19     while(~scanf("%d",&n)){
    20         while(n--){
    21             scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
    22 
    23             printf("%.3lf
    ",fabs(len(x2,y2)-len(x1,y1)));
    24         }
    25     }
    26 
    27 }

    2076  夹角有多大(题目已修改,注意读题)

     1 #include<stdio.h>
     2 #include<math.h>
     3 
     4 int main(){
     5     int n;
     6     double h,m,s,anh,anm,an;
     7     while(~scanf("%d",&n)){
     8         while(n--){
     9             scanf("%lf %lf %lf",&h,&m,&s);
    10             if(h>11){h = h-12;}
    11             anh = (h+(m+s/60)/60)*30;
    12             anm = (m+s/60)*6;
    13             an = fabs(anh - anm);
    14             if(an>180){an = 360 -an;}
    15             printf("%d
    ",(int)an);
    16         }
    17     }
    18 
    19 }

    2077  汉诺塔IV

     1 #include<stdio.h>
     2 #include<math.h>
     3 
     4 int main(){
     5     int n,i,m;
     6     _int64 s[21];
     7     s[0] = 0;
     8     s[1] = 1;
     9     for(i=2;i<21;i++){
    10         s[i] = s[i-1]*3 + 1;
    11     }
    12     while(~scanf("%d",&n)){
    13         while(n--){
    14             scanf("%d",&m);
    15             printf("%I64d
    ",s[m-1]*2+2);
    16         }
    17     }
    18     
    19 }

    2078  复习时间

     1 #include<stdio.h>
     2 #include<math.h>
     3 
     4 int main(){
     5     int t,n,m,s[45],i,j,temp,min;
     6     while(~scanf("%d",&t)){
     7         while(t--){
     8             scanf("%d %d",&n,&m);
     9             for(i=1;i<=n;i++){
    10                 scanf("%d",&s[i]);
    11             }
    12             for(i=1;i<=n;i++){
    13                 for(j=i+1;j<=n;j++){
    14                     if(s[i]<s[j]){
    15                         temp = s[i];
    16                         s[i] = s[j];
    17                         s[j] = temp;
    18                     }
    19                 }
    20             }
    21             
    22             printf("%d
    ",(100-s[n])*(100-s[n]));
    23         }
    24     }
    25 }

    2080  夹角有多大II

     1 #include<stdio.h>
     2 #include<math.h>
     3 #define PI 3.1415926
     4 
     5 int main(){
     6     int n;
     7     double x1,x2,y1,y2,m,t;
     8     while(~scanf("%d",&n)){
     9         while(n--){
    10             scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);
    11             m = x1*x2 + y1*y2;
    12             t = sqrt((x1*x1+y1*y1)*(x2*x2+y2*y2));
    13             printf("%.2lf
    ",acos(m/t)/PI*180);
    14         }
    15     }
    16 }

    抄公式就完事了

    2083   简易版之最短距离

     1 #include<stdio.h>
     2 #include<math.h>
     3 
     4 int main(){
     5     int n,m,i,j,s[501],temp,d;
     6     while(~scanf("%d",&n)){
     7         while(n--){
     8             scanf("%d",&m);
     9             for(i=0;i<m;i++){
    10                 scanf("%d",&s[i]);
    11             }
    12             for(i=0;i<m;i++){
    13                 for(j=i;j<m;j++){
    14                     if(s[i]>s[j]){
    15                         temp = s[i];
    16                         s[i] = s[j];
    17                         s[j] = temp;
    18                     }
    19                 }
    20             }
    21             d = 0;
    22             for(i=0;i<m;i++){
    23                 d += abs(s[i] - s[m/2]);
    24             }
    25             printf("%d
    ",d);
    26         }
    27     }
    28 }

    2084   数塔

     1 #include<stdio.h>
     2 int max(int a,int b){
     3     if(a>b){return a;}
     4     else{return b;}
     5 }
     6 
     7 int main(){
     8     int s[101][101];
     9     int dp[101][101];
    10     int c,n,i,j;
    11     while(~scanf("%d",&c)){
    12         while(c--){
    13             scanf("%d",&n);
    14             for(i=1;i<=n;i++){
    15                 for(j=1;j<=i;j++){
    16                     scanf("%d",&s[i][j]);
    17                 }
    18             }
    19             for(i=n;i>0;i--){
    20                 for(j=1;j<=i;j++){
    21                     if(i==n){
    22                         dp[i][j] = s[i][j];
    23                     }else{
    24                         dp[i][j] = max(s[i][j]+dp[i+1][j],s[i][j]+dp[i+1][j+1]);
    25                     }
    26                 }
    27             }
    28             printf("%d
    ",dp[1][1]);
    29         }
    30     }
    31 }

    2085   核反应堆

     1 #include<stdio.h>
     2 
     3 int main(){
     4     int n,i;
     5     _int64 h[35],l[35];
     6     h[0] = 1;l[0]=0;
     7     while(~scanf("%d",&n)){
     8         if(n==-1){break;}
     9         for(i=1;i<=n;i++){
    10             h[i] = h[i-1]*3 + l[i-1]*2;
    11             l[i] = h[i-1] + l[i-1];
    12         }
    13         printf("%I64d, %I64d
    ",h[n],l[n]);
    14     }
    15 
    16 }
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  • 原文地址:https://www.cnblogs.com/Ragd0ll/p/10514083.html
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