Problem A Coder
题意：水题
代码如下：
1 //2014-01-20-23.26 2 #include <iostream> 3 #include <cstdio> 4 #include <cstdlib> 5 #include <cmath> 6 #include <cstring> 7 #include <algorithm> 8 #include <queue> 9 #include <stack> 10 #include <vector> 11 #define MP(a, b) make_pair(a, b) 12 #define PB(a) push_back(a) 13 14 using namespace std; 15 16 typedef long long ll; 17 typedef pair<int ,int> pii; 18 typedef pair<unsigned int, unsigned int> puu; 19 typedef pair<int ,double> pid; 20 typedef pair<ll, int> pli; 21 22 const int INF = 0x3f3f3f3f; 23 const double eps = 1e-6; 24 const int LEN = 1010; 25 int Map[LEN][LEN]; 26 int ans; 27 28 int main() 29 { 30 // freopen("in.txt", "r", stdin); 31 32 int n; 33 while(scanf("%d", &n)!=EOF){ 34 memset(Map, 0, sizeof Map); 35 ans = 0; 36 for(int i=1; i<=n; i++){ 37 int f = i%2; 38 for(int j=1; j<=n; j++){ 39 Map[i][j] = f; 40 ans += f; 41 f = !f; 42 } 43 } 44 printf("%d\n", ans); 45 for(int i=1; i<=n; i++){ 46 for(int j=1; j<=n; j++){ 47 if(Map[i][j])printf("C"); 48 else printf("."); 49 }printf("\n"); 50 } 51 printf("\n"); 52 } 53 return 0; 54 }
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：
1 //2014-01-20-23.26 2 #include <iostream> 3 #include <cstdio> 4 #include <cstdlib> 5 #include <cmath> 6 #include <cstring> 7 #include <algorithm> 8 #include <queue> 9 #include <stack> 10 #include <vector> 11 #define MP(a, b) make_pair(a, b) 12 #define PB(a) push_back(a) 13 14 using namespace std; 15 16 typedef long long ll; 17 typedef pair<int ,int> pii; 18 typedef pair<unsigned int, unsigned int> puu; 19 typedef pair<int ,double> pid; 20 typedef pair<ll, int> pli; 21 22 const int INF = 0x3f3f3f3f; 23 const double eps = 1e-6; 24 const int LEN = 1010; 25 int ary[LEN][LEN], n, m; 26 27 int main() 28 { 29 // freopen("in.txt", "r", stdin); 30 31 int order; 32 33 while(scanf("%d%d%d", &n, &m, &order)!=EOF){ 34 for(int i=0; i<n; i++){ 35 for(int j=0; j<m; j++){ 36 scanf("%d", &ary[i][j]); 37 } 38 } 39 int ans = 0; 40 queue<pii> q; 41 for(int i=m-1; i>=0; i--){ 42 for(int j=0; j<i; j++){ 43 int f = 0; 44 for(int k=0; k<n; k++){ 45 if(order){ 46 if(ary[k][j]<ary[k][j+1]){ 47 swap(ary[k][j], ary[k][j+1]); 48 f = 1; 49 } 50 }else { 51 if(ary[k][j]>ary[k][j+1]){ 52 swap(ary[k][j], ary[k][j+1]); 53 f = 1; 54 } 55 } 56 } 57 if(f) { 58 if(order==0)q.push(MP(j,j+1)); 59 else q.push(MP(j+1,j)); 60 } 61 ans+=f; 62 } 63 } 64 printf("%d\n", ans); 65 while(!q.empty()){ 66 pii nv = q.front(); q.pop(); 67 printf("%d %d\n", nv.first+1, nv.second+1); 68 } 69 } 70 return 0; 71 }
高神的正确姿势：
1 #include <iostream> 2 #include <sstream> 3 #include <iomanip> 4 #include <functional> 5 #include <algorithm> 6 #include <vector> 7 #include <string> 8 #include <list> 9 #include <queue> 10 #include <deque> 11 #include <stack> 12 #include <set> 13 #include <map> 14 #include <cstdio> 15 #include <cstdlib> 16 #include <cmath> 17 #include <cstring> 18 #include <climits> 19 using namespace std; 20 #define XINF INT_MAX 21 #define INF 0x3FFFFFFF 22 #define MP(X,Y) make_pair(X,Y) 23 #define PB(X,Y) push_back(X,Y) 24 #define REP(X,N) for(int X=0;X<N;X++) 25 typedef pair<int,int> pii; 26 typedef vector<int> vi; 27 typedef long long ll; 28 29 30 int a[1000][100]; 31 int main() 32 { 33 int n,m,k; 34 while(cin>>n>>m>>k) 35 { 36 for(int i = 0; i < n; i++) 37 for(int j = 0; j < m; j++) 38 cin>>a[i][j]; 39 //if(k==0) //ascending 40 cout<<m*(m-1)/2<<endl; 41 for(int i = 0; i < m; i++) 42 for(int j = 0; j < m-i-1; j++) 43 { 44 if(k==0) 45 cout<<j+1<<‘ ‘<<j+2<<endl; 46 else 47 cout<<j+2<<‘ ‘<<j+1<<endl; 48 } 49 } 50 return 0; 51 }
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：
1 //2014-01-20-23.26 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 #include <algorithm> 6 #include <queue> 7 #include <stack> 8 #include <vector> 9 #define MP(a, b) make_pair(a, b) 10 #define PB(a) push_back(a) 11 12 using namespace std; 13 14 typedef long long ll; 15 typedef pair<int ,int> pii; 16 typedef pair<unsigned int, unsigned int> puu; 17 typedef pair<int ,double> pid; 18 typedef pair<ll, int> pli; 19 20 const int INF = 0x3f3f3f3f; 21 const double eps = 1e-6; 22 const int LEN = 201010; 23 24 int main() 25 { 26 // freopen("in.txt", "r", stdin); 27 28 int n, tt; 29 ll sum, ans; 30 while(scanf("%d",&n)!=EOF){ 31 ans = sum = 0; 32 for(int i=1; i<=n; i++){ 33 scanf("%d", &tt); 34 if(tt)sum += tt; 35 else ans+=sum; 36 } 37 printf("%I64d\n", ans); 38 } 39 return 0; 40 }