Codeforces Round #224 (Div. 2) 解题报告

Problem A Coder

题意：水题

代码如下：

//2014-01-20-23.26
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#define MP(a, b) make_pair(a, b)
#define PB(a) push_back(a)

using namespace std;

typedef long long ll;
typedef pair<int ,int> pii;
typedef pair<unsigned int, unsigned int> puu;
typedef pair<int ,double> pid;
typedef pair<ll, int> pli;

const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
const int LEN = 1010;
int Map[LEN][LEN];
int ans;

int main()
{
//    freopen("in.txt", "r", stdin);

    int n;
    while(scanf("%d", &n)!=EOF){
        memset(Map, 0, sizeof Map);
        ans = 0;
        for(int i=1; i<=n; i++){
            int f = i%2;
            for(int j=1; j<=n; j++){
                Map[i][j] = f;
                ans += f;
                f = !f;
            }
        }
        printf("%d
", ans);
        for(int i=1; i<=n; i++){
            for(int j=1; j<=n; j++){
                if(Map[i][j])printf("C");
                else printf(".");
            }printf("
");
        }
        printf("
");
    }
    return 0;
}

Problem B Multitasking

题意：对n个数组排序 每个长度都为n，让你输出一个序列（i，j）当i>j时才交换每个数组中的i,j号元素。

思路：这道题我十分二的，用了冒泡排序还去判断什么的。其实只要不管怎么样都交换m*(m-1)/2次就好了。

二B代码：

//2014-01-20-23.26
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#define MP(a, b) make_pair(a, b)
#define PB(a) push_back(a)

using namespace std;

typedef long long ll;
typedef pair<int ,int> pii;
typedef pair<unsigned int, unsigned int> puu;
typedef pair<int ,double> pid;
typedef pair<ll, int> pli;

const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
const int LEN = 1010;
int ary[LEN][LEN], n, m;

int main()
{
//    freopen("in.txt", "r", stdin);

    int order;

    while(scanf("%d%d%d", &n, &m, &order)!=EOF){
        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                scanf("%d", &ary[i][j]);
            }
        }
        int ans = 0;
        queue<pii> q;
        for(int i=m-1; i>=0; i--){
            for(int j=0; j<i; j++){
                int f = 0;
                for(int k=0; k<n; k++){
                    if(order){
                        if(ary[k][j]<ary[k][j+1]){
                            swap(ary[k][j], ary[k][j+1]);
                            f = 1;
                        }
                    }else {
                        if(ary[k][j]>ary[k][j+1]){
                            swap(ary[k][j], ary[k][j+1]);
                            f = 1;
                        }
                    }
                }
                if(f) {
                    if(order==0)q.push(MP(j,j+1));
                    else q.push(MP(j+1,j));
                }
                ans+=f;
            }
        }
        printf("%d
", ans);
        while(!q.empty()){
            pii nv = q.front(); q.pop();
            printf("%d %d
", nv.first+1, nv.second+1);
        }
    }
    return 0;
}

高神的正确姿势：

#include <iostream>
#include <sstream>
#include <iomanip>
#include <functional>
#include <algorithm>
#include <vector>
#include <string>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <climits>
using namespace std;
#define XINF INT_MAX
#define INF 0x3FFFFFFF
#define MP(X,Y) make_pair(X,Y)
#define PB(X,Y) push_back(X,Y)
#define REP(X,N) for(int X=0;X<N;X++)
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;


int a[1000][100];
int main()
{
    int n,m,k;
    while(cin>>n>>m>>k)
    {
        for(int i = 0; i < n; i++)
        for(int j = 0; j < m; j++)
            cin>>a[i][j];
        //if(k==0) //ascending
        cout<<m*(m-1)/2<<endl;
        for(int i = 0; i < m; i++)
        for(int j = 0; j < m-i-1; j++)
        {
            if(k==0)
                cout<<j+1<<' '<<j+2<<endl;
            else
                cout<<j+2<<' '<<j+1<<endl;
        }
    }
    return 0;
}

Problem C Milking cows

题意：一列牛1为头向右看的0为头向左看的。在一个牛被喂奶的时候有没喂过奶的牛看见会浪费一单元奶。问最少浪费多少单元奶？

思路：贪心先把朝一个方向的牛喂光，这样剩下来的就没有花费了。

代码如下：

//2014-01-20-23.26
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#define MP(a, b) make_pair(a, b)
#define PB(a) push_back(a)

using namespace std;

typedef long long ll;
typedef pair<int ,int> pii;
typedef pair<unsigned int, unsigned int> puu;
typedef pair<int ,double> pid;
typedef pair<ll, int> pli;

const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
const int LEN = 201010;

int main()
{
//    freopen("in.txt", "r", stdin);

    int n, tt;
    ll sum, ans;
    while(scanf("%d",&n)!=EOF){
        ans = sum = 0;
        for(int i=1; i<=n; i++){
            scanf("%d", &tt);
            if(tt)sum += tt;
            else ans+=sum;
        }
        printf("%I64d
", ans);
    }
    return 0;
}