跳过正文
  1. Posts/

codeforces #332 div 2 A. Patrick and Shopping

·2 分钟
 1#include <cstdio>
 2#include <iostream>
 3#include <cmath>
 4using namespace std;
 5long long d1,d2,d3;
 6int main()
 7{
 8cin>>d1>>d2>>d3;
 9long long ans = 999999999999;
10ans = min(ans,d1+d2+d3);
11ans = min (ans,d1*2+d2*2);
12ans = min (ans,d1*2+2*d3);
13ans = min(ans,d2*2+2*d3);
14cout<<ans<<endl;
15return 0;
16}

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a _d_1 meter long road between his house and the first shop and a _d_2 meter long road between his house and the second shop. Also, there is a road of length _d_3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.

Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn’t mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled.

Input

The first line of the input contains three integers _d_1, _d_2, _d_3 (1 ≤ _d_1, _d_2, _d_3 ≤ 108) – the lengths of the paths.

  • _d_1 is the length of the path connecting Patrick’s house and the first shop;
  • _d_2 is the length of the path connecting Patrick’s house and the second shop;
  • _d_3 is the length of the path connecting both shops.

Output

Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.

Sample test(s)

input

10 20 30

output

60

input

1 1 5

output

4

Note

The first sample is shown on the picture in the problem statement. One of the optimal routes is: house first shop second shop house.

In the second sample one of the optimal routes is: house first shop house second shop house.

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a _d_1 meter long road between his house and the first shop and a _d_2 meter long road between his house and the second shop. Also, there is a road of length _d_3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.

Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn’t mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled.

Input

The first line of the input contains three integers _d_1, _d_2, _d_3 (1 ≤ _d_1, _d_2, _d_3 ≤ 108) – the lengths of the paths.

  • _d_1 is the length of the path connecting Patrick’s house and the first shop;
  • _d_2 is the length of the path connecting Patrick’s house and the second shop;
  • _d_3 is the length of the path connecting both shops.

Output

Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.

Sample test(s)

input

10 20 30

output

60

input

1 1 5

output

4

Note

The first sample is shown on the picture in the problem statement. One of the optimal routes is: house first shop second shop house.

In the second sample one of the optimal routes is: house first shop house second shop house.

水题。

相关文章

幻方....

·1 分钟
c语言上机。。。。 c写的幻方。 1/************************************************************************* 2> File Name: code/class/7.c 3> Author: 111qqz 4> Email: rkz2013@126.com 5> Created Time: 2015年11月11日 星期三 19时31分50秒 6************************************************************************/ 7 8#include<stdio.h> 9#include <string.h> 10 11int n; 12int a[105][105]; 13 13 14 15void swap(int *a,int *b) 16{ 17int tmp; 18tmp = *a; 19*a = *b; 20*b = tmp; 21} 22int fix_x( int x,int k,int n) 23{ 24if (k%2==1) 25{ 26if (x==0) 27return n; 28else return x; 29} 30else 31{ 32if (x==n) 33return n+n; 34else return x; 35} 36} 37int fix_y ( int y,int k,int n) 38{ 39if (k<3) 40{ 41if (y==n+1) 42return 1; 43else return y; 44} 45else 46{ 47if (y==2*n+1) 48return n+1; 49else return y; 50} 51// if (y==n+1) 52// return 1; 53// else return y; 54} 55void print() 56{ 57for ( int i = 1 ; i <= n ; i++) 58{ 59for ( int j = 1 ; j <= n ; j++) 60printf("%d ",a[i][j]); 61 61 62printf("n"); 63} 64 64 65} 66 66 67void OddMagic(int n,int x,int y,int k) //k表示4中状态。。。。 68{ 69 69 70int cur ; 71if (k==1) cur = 1; 72if (k==4) cur = n*n+1; 73if (k==3) cur = n*n*2+1; 74if (k==2) cur = n*n*3+1; 75int cnt = 1; 76while (cnt<=n*n) 77{ 78a[x][y]=cur; 79int prex = x; 80int prey = y; 81cur++; 82cnt++; 83x--; 84y++; 85x = fix_x(x,k,n); 86y = fix_y(y,k,n); 87if (a[x][y]) 88{ 89x = prex+1; 90y = prey; 91} 92 92 93} 94 94 95} 96int main() 97{ 98memset(a,sizeof(a),0); 99scanf("%d",&n); 100if (n%2==1) 101{ 102int x = 1; 103int y = n/2+1; 104OddMagic(n,x,y,1); 105} 106else 107{ 108if (n%4==0) 109{ 110for ( int i = 1,num=1 ; i <= n ; i++) 111for ( int j = 1 ; j <= n ; j++,num++) 112a[i][j]=num; 113 114 115for ( int i = 1 ; i <= n ; i++) 116{ 117for ( int j = 1 ; j <= n ; j++) 118{ 119if (i==j||i+j>=n+1) continue; 120int tmp; 121tmp = a[i][j]; 122a[i][j] = a[n+1-i][n+1-j]; 123a[n+1-i][n+1-j] = tmp; 124} 125} 126} 127else 128{ 129int x = 1; 130int y = n/4+1; 131int hn = n/2; 132 133OddMagic(hn,x,y,1); 134OddMagic(hn,x+hn,y,2); 135OddMagic(hn,x,y+hn,3); 136OddMagic(hn,x+hn,y+hn,4); 137 138int m = n/4; 139for ( int i = 1 ; i <= hn ;i++) 140{ 141for ( int j = 1 ; j <= m ; j++) 142{ 143int tmp; 144if (i==m+1&&j;==m) 145{ 146tmp = a[m+1][m+1]; 147a[m+1][m+1] = a[m+1+hn][m+1]; 148a[m+1+hn][m+1] = tmp; 149continue; 150 151} 152tmp = a[i][j]; 153a[i][j] = a[i+hn][j]; 154a[i+hn][j] = tmp; 155// swap(a[i][j],a[i+n][j]); 156} 157} 158 159for ( int i = 1 ; i <= hn ; i++) 160{ 161for ( int j = n ; j>=n-m+2 ; j--) 162{ 163int tmp; 164tmp = a[i][j]; 165a[i][j] = a[i+hn][j]; 166a[i+hn][j] = tmp; 167} 168} 169 170 171 172 173} 174} 175print(); 176 177}