题目链接：http://codeforces.com/problemset/problem/93/B
Students love to celebrate their holidays. Especially if the holiday is the day of the end of exams!
Despite the fact that Igor K., unlike his groupmates, failed to pass a programming test, he decided to invite them to go to a cafe so that each of them could drink a bottle of... fresh cow milk. Having entered the cafe, the m friends found n different kinds of milk on the menu, that's why they ordered n bottles — one bottle of each kind. We know that the volume of milk in each bottle equals w.
When the bottles were brought in, they decided to pour all the milk evenly among the m cups, so that each got a cup. As a punishment for not passing the test Igor was appointed the person to pour the milk. He protested that he was afraid to mix something up and suggested to distribute the drink so that the milk from each bottle was in no more than two different cups. His friends agreed but they suddenly faced the following problem — and what is actually the way to do it?
Help them and write the program that will help to distribute the milk among the cups and drink it as quickly as possible!
Note that due to Igor K.'s perfectly accurate eye and unswerving hands, he can pour any fractional amount of milk from any bottle to any cup.
The only input data file contains three integers n, w and m (1 ≤ n ≤ 50, 100 ≤ w ≤ 1000, 2 ≤ m ≤ 50), where n stands for the number of ordered bottles, w stands for the volume of each of them and m stands for the number of friends in the company.
Print on the first line "YES" if it is possible to pour the milk so that the milk from each bottle was in no more than two different cups. If there's no solution, print "NO".
If there is a solution, then print m more lines, where the i-th of them describes the content of the i-th student's cup. The line should consist of one or more pairs that would look like "b v". Each such pair means that v (v > 0) units of milk were poured into the i-th cup from bottle b (1 ≤ b ≤ n). All numbers b on each line should be different.
If there are several variants to solve the problem, print any of them. Print the real numbers with no less than 6 digits after the decimal point.
2 500 3
YES
1 333.333333
2 333.333333
2 166.666667 1 166.666667
4 100 5
YES
3 20.000000 4 60.000000
1 80.000000
4 40.000000 2 40.000000
3 80.000000
2 60.000000 1 20.000000
4 100 7
NO
5 500 2
YES
4 250.000000 5 500.000000 2 500.000000
3 500.000000 1 500.000000 4 250.000000
题目大意：输入n,w,m 分别代表n杯不同的饮料，每杯饮料有w体积，要平均分为m杯
思路：这题A出来感觉非常不简单，一天就做了这一道题。。。 其实这题要懂的就是如何贪心，然后如何存数据答案，还有一个很重要的就是double的精度问题，控制不好就别想过了
贪心思想是： 每次都使得当前用的饮料尽可能的多，如果用了两次都没有用完的话，那就显然是“NO”了，贪心思想很简单。
然后关于存储数据的:因为我们并不知道装满一杯要多少种饮料，所以这个就必须要用动态存储了，这就需要用到vector,我们要存储的有第几杯，用了哪种饮料，用了多少，所以要三个
变量来存储。 学了一下如果用vector来存储三个变量，就是再用一个pair，具体怎么用在代码中可以看到
double精度的控制：这题本来自己感觉是没有问题的，然后提交,wa在第十个吧，忘记了，然后试了数据，发现结果跟自己想的不一样，后来改代码，把过程中的结果输出来，发现
在运行过程中double 保留的是三位小数，但是相减却又保留的是六位小数，比如一个是333.333，1000和它相减则会得到666.666667，，但是下一次和666.667比较，如果我们简单
的比较二者是否相等的话，显然是不相等的，但是二者实际上是相等的，所以要二者相减判断差值是否在误差之外就行了。
看代码
#include<iostream> #include<string.h> #include<map> #include<cstdio> #include<cstring> #include<stdio.h> #include<cmath> #include<ctype.h> #include<math.h> #include<algorithm> #include<set> #include<queue> typedef long long ll; using namespace std; const ll mod=1e9+7; const int maxn=1e3+10; const int maxk=5e3+10; const int maxx=1e4+10; const ll maxe=1000+10; #define INF 0x3f3f3f3f3f3f #define Lson l,mid,rt<<1 #define Rson mid+1,r,rt<<1|1 typedef pair<int,double> p; int main() { //vector<pair<int,double> > ans[100]; vector<p>ans[100]; p fun; int n,w,m; cin>>n>>w>>m; if(m>n*2) cout<<"NO"<<endl;//n瓶饮料最多能分出2*n杯饮料，所以大于它直接NO else { int cnt=1,sum1=1,sum3=0;//cnt代表第几瓶，sum1代表第几个人,sum3判断是否用了两次以上 double ave=1.0*n*w/m; double sum=ave,sum2=w;//补满一瓶还需的容量,一瓶还剩多少 while(cnt<=n&&sum1<=m)// { if(fabs(sum-sum2)<1e-6)//精度判断，一定要放在最前面，不然会判断别的，也就wa了 { //cout<<"2"<<endl; fun.first=cnt; fun.second=sum2; ans[sum1].push_back(fun); //ans[sum1].push_back({cnt,sum2}); sum3++; if(sum3>2) { cout<<"NO"<<endl; return 0; } //ans[cnt].insert(n); //ans.insert(cnt,n); sum3=0; sum2=w; sum=ave; cnt++; sum1++; //cout<<cnt<<" "<<sum<<" "<<sum2<<endl; } else if(sum>sum2) { // cout<<"1"<<endl; fun.first=cnt; fun.second=sum2; ans[sum1].push_back(fun); //ans[sum1].push_back({cnt,sum2}); sum3++; if(sum3>2) { cout<<"NO"<<endl; return 0; } sum3=0; sum-=sum2; sum2=w; cnt++; } //else if(sum==sum2) else { //cout<<"3"<<endl; //printf("%.3lf %.6lf ",sum,sum2); fun.first=cnt; fun.second=sum; ans[sum1].push_back(fun); //ans[sum1].push_back({cnt,sum}); sum2=sum2-sum; if(sum2==0.000000) cout<<"sss"<<endl; sum=ave; sum3++; sum1++; } if(sum3>2) { cout<<"NO"<<endl; return 0; } } cout<<"YES"<<endl; for(int i=1;i<=m;i++) { for(int j=0;j<ans[i].size();j++) { printf("%d %.6lf ",ans[i][j].first,ans[i][j].second); //cout<<ans[i][j].first<<" "<<ans[i][j].second; } cout<<endl; } } return 0; }