POJ 3312 Mahershalalhashbaz, Nebuchadnezzar, and Billy Bob Benjamin Go to the Regionals (水题，贪心)

题意：给定 n 个字符串，一个k，让你把它们分成每组k个，要保证每组中每个字符串长度与它们之和相差不能超2.

析：贪心策略就是长度相差最小的放上块。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <functional>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <deque>
#include <map>
#include <cctype>
#include <stack>
#include <sstream>
using namespace std ;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1000 + 5;
const int mod = 1e9 + 7;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
int n, m;
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
char s[maxn];
vector<int> v;

int main(){
    int kase = 0;
    while(scanf("%d %d", &n, &m) == 2){
        if(!m && !n)  break;

        v.clear();
        for(int i = 0; i < n; ++i){
            scanf("%s", s);
            v.push_back(strlen(s));
        }
        sort(v.begin(), v.end());
        bool ok = true;
        for(int i = 0; i < n/m; i++){
            double cnt = 0.0;
            for(int j = 0; j < m; ++j)
                cnt += v[i*m+j];
            cnt /= (double)m*1.0;
            for(int j = 0; j < m; ++j)
                if(abs((double)v[i*m+j]*1.0 - cnt) > 2.0){ ok = false; break; }
        }
        if(kase++)  printf("\n");
        printf("Case %d: %s\n", kase, ok ? "yes" : "no");
    }
    return 0;
}