POJ 2418-Hardwood Species(map)
| Time Limit: 10000MS | Memory Limit: 65536K | |
| Total Submissions: 18770 | Accepted: 7405 |
Description
America's temperate climates produce forests with hundreds of hardwood species -- trees that share certain biological characteristics. Although oak, maple and cherry all are types of hardwood trees, for example, they are different species. Together, all the
hardwood species represent 40 percent of the trees in the United States.
On the other hand, softwoods, or conifers, from the Latin word meaning "cone-bearing," have needles. Widely available US softwoods include cedar, fir, hemlock, pine, redwood, spruce and cypress. In a home, the softwoods are used primarily as structural lumber
such as 2x4s and 2x6s, with some limited decorative applications.
Using satellite imaging technology, the Department of Natural Resources has compiled an inventory of every tree standing on a particular day. You are to compute the total fraction of the tree population represented by each species.
Input
Output
Sample Input
Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow
Sample Output
Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.3448
Willow 3.4483
Yellow Birch 3.4483
题意:给一堆单词(有反复),然后按字典序打印出全部单词(无反复) 而且输出每一个单词占总单词数的百分率。
值得注意的是G++提交要用%f 而C++用%lf
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cctype>
#include <cstdlib>
#include <algorithm>
#include <set>
#include <vector>
#include <string>
#include <map>
#include <queue>
using namespace std;
int main()
{
ios::sync_with_stdio(false);
string x;
long long cnt=0;
map <string,int> m;
while(getline(cin,x)){
cnt++;
m[x]++; }
map <string,int>::iterator it=m.begin();
while(it!=m.end())
{
printf("%s ",it->first.c_str());
printf("%.4lf\n",it->second*100.0/cnt);
++it;
}
return 0;
}
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