标签：des style blog http io ar color os sp
哎，最近弱爆了，，，不过这题还是不错滴~~ 要考虑完整各种情况
| 8795058 | 2014-11-22 06:52:58 | njczy2010 | B - Candy Boxes | GNU C++ | Accepted | 31 ms | 4 KB |
| 8795016 | 2014-11-22 06:48:15 | njczy2010 | B - Candy Boxes | GNU C++ | Wrong answer on test 13 | 30 ms | 0 KB |
| 8795000 | 2014-11-22 06:44:39 | njczy2010 | B - Candy Boxes | GNU C++ | Wrong answer on test 5 | 15 ms | 0 KB |
There is an old tradition of keeping 4 boxes of candies in the house in Cyberland. The numbers of candies are special if their arithmetic mean, their median and their range are all equal. By definition, for a set {x1, x2, x3, x4} (x1 ≤ x2 ≤ x3 ≤ x4) arithmetic mean is 
, median is 
and range is x4 - x1. The arithmetic mean and median are not necessary integer. It is well-known that if those three numbers are same, boxes will create a "debugging field" and codes in the field will have no bugs.
For example, 1, 1, 3, 3 is the example of 4 numbers meeting the condition because their mean, median and range are all equal to 2.
Jeff has 4 special boxes of candies. However, something bad has happened! Some of the boxes could have been lost and now there are only n (0 ≤ n ≤ 4) boxes remaining. The i-th remaining box contains ai candies.
Now Jeff wants to know: is there a possible way to find the number of candies of the 4 - n missing boxes, meeting the condition above (the mean, median and range are equal)?
The first line of input contains an only integer n (0 ≤ n ≤ 4).
The next n lines contain integers ai, denoting the number of candies in the i-th box (1 ≤ ai ≤ 500).
In the first output line, print "YES" if a solution exists, or print "NO" if there is no solution.
If a solution exists, you should output 4 - n more lines, each line containing an integer b, denoting the number of candies in a missing box.
All your numbers b must satisfy inequality 1 ≤ b ≤ 106. It is guaranteed that if there exists a positive integer solution, you can always find such b‘s meeting the condition. If there are multiple answers, you are allowed to print any of them.
Given numbers ai may follow in any order in the input, not necessary in non-decreasing.
ai may have stood at any positions in the original set, not necessary on lowest n first positions.
2 1 1
YES 3 3
3 1 1 1
NO
4 1 2 2 3
YES
For the first sample, the numbers of candies in 4 boxes can be 1, 1, 3, 3. The arithmetic mean, the median and the range of them are all 2.
For the second sample, it‘s impossible to find the missing number of candies.
In the third example no box has been lost and numbers satisfy the condition.
You may output b in any order.
主要就是根据那三个等式，化简，得两个方程：
x4=3*x1;
x2+x3=4*x1;
1 #include<iostream> 2 #include<cstring> 3 #include<cstdlib> 4 #include<cstdio> 5 #include<algorithm> 6 #include<cmath> 7 #include<queue> 8 #include<map> 9 #include<set> 10 #include<string> 11 //#include<pair> 12 13 #define N 10005 14 #define M 1005 15 #define mod 1000000007 16 //#define p 10000007 17 #define mod2 1000000000 18 #define ll long long 19 #define LL long long 20 #define eps 1e-9 21 #define maxi(a,b) (a)>(b)? (a) : (b) 22 #define mini(a,b) (a)<(b)? (a) : (b) 23 24 using namespace std; 25 26 int n; 27 int a[10]; 28 int b[10]; 29 int flag; 30 31 void ini() 32 { 33 flag=0; 34 for(int i=1;i<=n;i++){ 35 scanf("%d",&a[i]); 36 } 37 sort(a+1,a+1+n); 38 } 39 40 void solve0() 41 { 42 flag=1; 43 b[1]=1;b[2]=1;b[3]=3;b[4]=3; 44 } 45 46 void solve1() 47 { 48 flag=1; 49 b[1]=a[1];b[2]=3*a[1];b[3]=3*a[1]; 50 } 51 52 void solve2() 53 { 54 if(a[2]%3==0){ 55 if(a[2]/3==a[1]){ 56 flag=1; 57 b[1]=a[1];b[2]=a[2]; 58 } 59 else if(a[2]/3<a[1]){ 60 flag=1;b[1]=a[2]/3;b[2]=a[2]+b[1]-a[1]; 61 } 62 else{ 63 return; 64 } 65 } 66 else{ 67 if(a[1]*3<=a[2]){ 68 return; 69 } 70 else{ 71 flag=1; 72 b[2]=a[1]*3; 73 b[1]=b[2]+a[1]-a[2]; 74 } 75 } 76 } 77 78 void solve3() 79 { 80 if(a[3]%3!=0){ 81 if(a[1]*3<=a[3]){ 82 return; 83 } 84 else{ 85 if(a[1]*4==a[2]+a[3]){ 86 flag=1;b[1]=a[1]*3; 87 } 88 else return; 89 } 90 } 91 else{ 92 if(a[3]/3==a[1]){ 93 flag=1; 94 b[1]=a[1]+a[3]-a[2]; 95 } 96 else if(a[3]/3<a[1]){ 97 b[1]=a[3]/3; 98 if(b[1]+a[3]==a[1]+a[2]){ 99 flag=1; 100 } 101 else return; 102 } 103 else{ 104 return; 105 } 106 } 107 } 108 109 void solve4() 110 { 111 if(a[4]==3*a[1] && a[2]+a[3]==4*a[1]){ 112 flag=1; 113 } 114 } 115 116 void solve() 117 { 118 if(n==4){ 119 solve4(); 120 } 121 else if(n==0){ 122 solve0(); 123 } 124 else if(n==1){ 125 solve1(); 126 } 127 else if(n==2){ 128 solve2(); 129 } 130 else if(n==3){ 131 solve3(); 132 } 133 } 134 135 void out() 136 { 137 int i; 138 if(flag==1){ 139 printf("YES\n"); 140 for(i=1;i<=4-n;i++){ 141 printf("%d\n",b[i]); 142 } 143 } 144 else{ 145 printf("NO\n"); 146 } 147 } 148 149 int main() 150 { 151 //freopen("data.in","r",stdin); 152 //freopen("data.out","w",stdout); 153 //scanf("%d",&T); 154 //for(int ccnt=1;ccnt<=T;ccnt++) 155 // while(T--) 156 while(scanf("%d",&n)!=EOF) 157 { 158 // if(n==0 && m==0 ) break; 159 //printf("Case %d: ",ccnt); 160 ini(); 161 solve(); 162 out(); 163 } 164 165 return 0; 166 }
Codeforces Round #278 (Div. 2) B. Candy Boxes [brute force+constructive algorithms]
标签：des style blog http io ar color os sp
原文地址：http://www.cnblogs.com/njczy2010/p/4114977.html
