开了个小号去做div2写一下解题报告。
Problem A Inna and Alarm Clock
简单题。直接开数组标记一下即可。代码如下：
1 /************************************************** 2 * Author : xiaohao Z 3 * Blog : http://www.cnblogs.com/shu-xiaohao/ 4 * Last modified : 2014-02-11 23:26 5 * Filename : Round_229_2_A.cpp 6 * Description : 7 * ************************************************/ 8 9 #include <iostream> 10 #include <cstdio> 11 #include <cstring> 12 #include <cstdlib> 13 #include <cmath> 14 #include <algorithm> 15 #include <queue> 16 #include <stack> 17 #include <vector> 18 #include <set> 19 #include <map> 20 #define MP(a, b) make_pair(a, b) 21 #define PB(a) push_back(a) 22 23 using namespace std; 24 typedef long long ll; 25 typedef pair<int, int> pii; 26 typedef pair<unsigned int,unsigned int> puu; 27 typedef pair<int, double> pid; 28 typedef pair<ll, int> pli; 29 typedef pair<int, ll> pil; 30 31 const int INF = 0x3f3f3f3f; 32 const double eps = 1E-6; 33 const int LEN = 101; 34 35 int main() 36 { 37 // freopen("in.txt", "r", stdin); 38 39 int n, a, b, col[LEN], row[LEN]; 40 while(scanf("%d", &n)!=EOF){ 41 memset(row, 0, sizeof row); 42 memset(col, 0, sizeof col); 43 for(int i=0; i<n; i++){ 44 scanf("%d%d", &a, &b); 45 col[b] = 1;row[a] = 1; 46 } 47 a = b = 0; 48 for(int i=0; i<LEN; i++){ 49 if(col[i]) a++; 50 if(row[i]) b++; 51 } 52 printf("%d ", min(a, b)); 53 } 54 return 0; 55 }
Problem B Inna, Dima and Song
思路：首先(b[i] > 1 && b[i] <= a[i]*2)才能满足能弹出来，否则-1.满足的话(ll)(b[i]/2)*(ll)(b[i]-(b[i]/2))即是最大之（注意我的强制类型转换否则会超或者直接用ll）
代码如下：
1 /************************************************** 2 * Author : xiaohao Z 3 * Blog : http://www.cnblogs.com/shu-xiaohao/ 4 * Last modified : 2014-02-11 23:28 5 * Filename : Round_229_2_B.cpp 6 * Description : 7 * ************************************************/ 8 9 #include <iostream> 10 #include <cstdio> 11 #include <cstring> 12 #include <cstdlib> 13 #include <cmath> 14 #include <algorithm> 15 #include <queue> 16 #include <stack> 17 #include <vector> 18 #include <set> 19 #include <map> 20 #define MP(a, b) make_pair(a, b) 21 #define PB(a) push_back(a) 22 23 using namespace std; 24 typedef long long ll; 25 typedef pair<int, int> pii; 26 typedef pair<unsigned int,unsigned int> puu; 27 typedef pair<int, double> pid; 28 typedef pair<ll, int> pli; 29 typedef pair<int, ll> pil; 30 31 const int INF = 0x3f3f3f3f; 32 const double eps = 1E-6; 33 const int LEN = 100010; 34 int a[LEN], b[LEN]; 35 36 int main() 37 { 38 // freopen("in.txt", "r", stdin); 39 40 int n; 41 while(scanf("%d", &n)!=EOF){ 42 for(int i=0; i<n; i++){ 43 scanf("%d", &a[i]); 44 } 45 for(int i=0; i<n; i++){ 46 scanf("%d", &b[i]); 47 } 48 ll ans = 0; 49 for(int i=0; i<n; i++){ 50 if(b[i] > 1 && b[i] <= a[i]*2)ans+=(ll)(b[i]/2)*(ll)(b[i]-(b[i]/2)); 51 else ans --; 52 } 53 printf("%I64d ", ans); 54 } 55 return 0; 56 }
Problem C Inna and Candy Boxes
题意：有一排盒子每个盒子里面两种情况有一个糖和没有。给长度n，整数k。然后m个区间查询要求把[l,r]中变成只有l+tk-1有糖需要几步操作。每一次操作能拿走一颗糖，放进一颗糖。
思路：利用k比较小的特征把k看成一个循环节，对k的余数分别递推dp[i]表示i之前和i模k相等的有几颗糖。在更具询问统计循环节上的每一个就行了。
代码如下：
1 /************************************************** 2 * Author : xiaohao Z 3 * Blog : http://www.cnblogs.com/shu-xiaohao/ 4 * Last modified : 2014-02-11 23:27 5 * Filename : Round_229_2_C.cpp 6 * Description : 7 * ************************************************/ 8 9 #include <iostream> 10 #include <cstdio> 11 #include <cstring> 12 #include <cstdlib> 13 #include <cmath> 14 #include <algorithm> 15 #include <queue> 16 #include <stack> 17 #include <vector> 18 #include <set> 19 #include <map> 20 #define MP(a, b) make_pair(a, b) 21 #define PB(a) push_back(a) 22 23 using namespace std; 24 typedef long long ll; 25 typedef pair<int, int> pii; 26 typedef pair<unsigned int,unsigned int> puu; 27 typedef pair<int, double> pid; 28 typedef pair<ll, int> pli; 29 typedef pair<int, ll> pil; 30 31 const int INF = 0x3f3f3f3f; 32 const double eps = 1E-6; 33 const int LEN = 100000+10; 34 int dp[LEN]; 35 char str[LEN]; 36 int n, k, m, l, r; 37 38 int main() 39 { 40 // freopen("in.txt", "r", stdin); 41 42 while(scanf("%d%d%d", &n, &k, &m)!=EOF){ 43 getchar(); 44 gets(str); 45 for(int i=0; i<k; i++) dp[i] = (str[i]=='1'?1:0); 46 for(int i=k; i<n; i++) dp[i] = dp[i-k]+(str[i]=='1'?1:0); 47 for(int i=0; i<m; i++){ 48 scanf("%d%d", &l, &r); 49 l--;r--; 50 int ans = 0; 51 for(int i=1; i<=k; i++){ 52 int hi, lo; 53 if(l-i < 0) lo = 0; 54 else lo = dp[l-i]; 55 if(r+1-i < 0) hi = 0; 56 else hi = dp[r+1-i]; 57 if(i==1) ans += (((r-l+1)/k) - (hi-lo)); 58 else ans += (hi-lo); 59 } 60 printf("%d ", ans); 61 } 62 } 63 return 0; 64 }
Problem D Inna and Sweet Matrix
题意：在一张地图上一开始全空。每次选一条路上没有糖的路径走过去放一颗糖。每格只能放一颗，有k颗问你放完的最小花费是多少？花费为走过的总路程。
思路：贪心思想。按bfs顺序确定放糖位置，然后每次放最远的糖，走过去按x先增y后增。具体参照我的排序函数。
代码如下：
1 /************************************************** 2 * Author : xiaohao Z 3 * Blog : http://www.cnblogs.com/shu-xiaohao/ 4 * Last modified : 2014-02-11 23:29 5 * Filename : Round_229_2_D.cpp 6 * Description : 7 * ************************************************/ 8 9 #include <iostream> 10 #include <cstdio> 11 #include <cstring> 12 #include <cstdlib> 13 #include <cmath> 14 #include <algorithm> 15 #include <queue> 16 #include <stack> 17 #include <vector> 18 #include <set> 19 #include <map> 20 #define MP(a, b) make_pair(a, b) 21 #define PB(a) push_back(a) 22 23 using namespace std; 24 typedef long long ll; 25 typedef pair<int, int> pii; 26 typedef pair<unsigned int,unsigned int> puu; 27 typedef pair<int, double> pid; 28 typedef pair<ll, int> pli; 29 typedef pair<int, ll> pil; 30 31 const int INF = 0x3f3f3f3f; 32 const double eps = 1E-6; 33 const int LEN = 55; 34 int Map[LEN][LEN], n, m; 35 struct P{int x, y;}; 36 struct cmp{ 37 bool operator() (P a, P b){ 38 if(a.x+a.y!=b.x+b.y) return a.x+a.y < b.x+b.y; 39 else return a.x > b.x; 40 } 41 }; 42 priority_queue<P, vector<P>, cmp> pq; 43 P MPP(int a, int b){ 44 P ret; 45 ret.x = a; 46 ret.y = b; 47 return ret; 48 } 49 int xx[] = {0, 0, 1,-1}; 50 int yy[] = {1,-1, 0, 0}; 51 52 53 int put(int num){ 54 int ret = 1; 55 queue<pii> q; 56 q.push(MP(0,0)); 57 Map[0][0] = 1; 58 pq.push(MPP(0,0)); 59 num--; if(num == 0) return ret; 60 while(!q.empty()){ 61 pii nv = q.front(); q.pop(); 62 for(int i=0; i<4; i++){ 63 int tx = nv.first+xx[i]; 64 int ty = nv.second+yy[i]; 65 if(tx >= 0 && tx < n && ty >= 0 && ty < m && !Map[tx][ty]){ 66 Map[tx][ty] = 1; 67 q.push(MP(tx, ty)); 68 pq.push(MPP(tx, ty)); 69 num--; 70 ret += (tx+ty+1); 71 if(num == 0) return ret; 72 } 73 } 74 } 75 } 76 77 int main() 78 { 79 // freopen("in.txt", "r", stdin); 80 81 int st; 82 while(scanf("%d%d%d", &n, &m, &st)!=EOF){ 83 memset(Map, 0, sizeof Map); 84 while(!pq.empty())pq.pop(); 85 int ans = put(st); 86 printf("%d ", ans); 87 while(!pq.empty()){ 88 P nv = pq.top();pq.pop(); 89 nv.x++,nv.y++; 90 int x = 1, y = 1; 91 while(1){ 92 if(x == nv.x && y == nv.y){ 93 printf("(%d,%d) ", nv.x, nv.y); 94 break; 95 } 96 printf("(%d,%d) ", x, y); 97 if(x < nv.x) x++; 98 else if(y < nv.y) y++; 99 } 100 } 101 } 102 return 0; 103 }