CodeForces 221(div 2)

A

无trick水题。。。

 1 /*
 2  * Author:  Plumrain
 3  * Created Time:  2013-12-24 22:26
 4  * File Name: B.cpp
 5  */
 6 #include <iostream>
 7 #include <cstdio>
 8 #include <cstring>
 9 #include <string>
10 #include <cmath>
11 #include <algorithm>
12 #include <vector>
13 #include <cstdlib>
14 #include <sstream>
15 #include <fstream>
16 #include <list>
17 #include <deque>
18 #include <queue>
19 #include <stack>
20 #include <map>
21 #include <set>
22 #include <bitset>
23 #include <cctype>
24 #include <ctime>
25 #include <utility>
26 
27 using namespace std;
28 
29 #define clr0(x) memset(x, 0, sizeof(x))
30 #define clr1(x) memset(x, -1, sizeof(x))
31 #define pb push_back
32 #define sz(v) ((int)(v).size())
33 #define all(t) t.begin(),t.end()
34 #define INF 999999999999999999
35 #define zero(x) (((x)>0?(x):-(x))<eps)
36 #define out(x) cout<<#x<<":"<<(x)<<endl
37 #define tst(a) cout<<a<<" "
38 #define tst1(a) cout<<#a<<endl
39 #define CINBEQUICKER std::ios::sync_with_stdio(false)
40 
41 const double eps = 1e-8;
42 const double PI = atan(1.0)*4;
43 const int inf = 2147483647 / 2;
44 
45 typedef vector<int> vi;
46 typedef vector<string> vs;
47 typedef vector<double> vd;
48 typedef pair<int, int> pii;
49 typedef long long int64;
50 
51 inline int Mymod (int a, int b) {int x=a%b; if(x<0) x+=b; return x;}
52 
53 int ru[105], chu[105];
54 
55 int main()
56 {
57 //    freopen("a.in","r",stdin);
58 //    freopen("a.out","w",stdout);
59 //    std::ios::sync_with_stdio(false);
60     int n, m;
61     while (scanf ("%d%d", &n , &m) != EOF){
62         int t1, t2, w;
63         clr0 (chu); clr0 (ru);
64         for (int i = 0; i < m; ++ i){
65             scanf ("%d%d%d", &t1, &t2, &w);
66             chu[--t1] += w;
67             ru[--t2] += w;
68         }
69         int ans = 0;
70         for (int i = 0; i < n; ++ i)
71             ans += abs(chu[i] - ru[i]);
72         printf ("%d\n", ans / 2);
73     }
74     return 0;
75 }
View Code

 

B

YY题。。。结论直接看代码就好了。。。比赛的时候YY出了结论算样例算错了，然后就去想别的方法了。。。。。。

 1 /*
 2  * Author:  Plumrain
 3  * Created Time:  2013-12-24 22:26
 4  * File Name: B.cpp
 5  */
 6 #include <iostream>
 7 #include <cstdio>
 8 #include <cstring>
 9 #include <string>
10 #include <cmath>
11 #include <algorithm>
12 #include <vector>
13 #include <cstdlib>
14 #include <sstream>
15 #include <fstream>
16 #include <list>
17 #include <deque>
18 #include <queue>
19 #include <stack>
20 #include <map>
21 #include <set>
22 #include <bitset>
23 #include <cctype>
24 #include <ctime>
25 #include <utility>
26 
27 using namespace std;
28 
29 #define clr0(x) memset(x, 0, sizeof(x))
30 #define clr1(x) memset(x, -1, sizeof(x))
31 #define pb push_back
32 #define sz(v) ((int)(v).size())
33 #define all(t) t.begin(),t.end()
34 #define INF 999999999999999999
35 #define zero(x) (((x)>0?(x):-(x))<eps)
36 #define out(x) cout<<#x<<":"<<(x)<<endl
37 #define tst(a) cout<<a<<" "
38 #define tst1(a) cout<<#a<<endl
39 #define CINBEQUICKER std::ios::sync_with_stdio(false)
40 
41 const double eps = 1e-8;
42 const double PI = atan(1.0)*4;
43 const int inf = 2147483647 / 2;
44 
45 typedef vector<int> vi;
46 typedef vector<string> vs;
47 typedef vector<double> vd;
48 typedef pair<int, int> pii;
49 typedef long long int64;
50 
51 inline int Mymod (int a, int b) {int x=a%b; if(x<0) x+=b; return x;}
52 
53 int ru[105], chu[105];
54 
55 int main()
56 {
57 //    freopen("a.in","r",stdin);
58 //    freopen("a.out","w",stdout);
59 //    std::ios::sync_with_stdio(false);
60     int n, m;
61     while (scanf ("%d%d", &n , &m) != EOF){
62         int t1, t2, w;
63         clr0 (chu); clr0 (ru);
64         for (int i = 0; i < m; ++ i){
65             scanf ("%d%d%d", &t1, &t2, &w);
66             chu[--t1] += w;
67             ru[--t2] += w;
68         }
69         int ans = 0;
70         for (int i = 0; i < n; ++ i)
71             ans += abs(chu[i] - ru[i]);
72         printf ("%d\n", ans / 2);
73     }
74     return 0;
75 }
View Code

 

C

题意：给一个很大的数m，这个数的各个位上的数字中一定含有至少1个1,6,8,9。你可以重新组织所有数字的顺序，使得重新排列之后的数能被7整除。10^4 <= m <= 10^(10^6)。

　　　注意，排列之后的数不能含有前导0。

解法：把1,6,8,9四个数字放在最后面，根据前面的数*10000除以7的余数，来决定1,6,8,9四个数字的排列顺序即可。至于有没有前导0，特殊处理一下即可。

tag:math, think

  1 /*
  2  * Author:  Plumrain
  3  * Created Time:  2013-12-25 14:09
  4  * File Name: C.cpp
  5  */
  6 #include <iostream>
  7 #include <cstdio>
  8 #include <cstring>
  9 #include <string>
 10 #include <cmath>
 11 #include <algorithm>
 12 #include <vector>
 13 #include <cstdlib>
 14 #include <sstream>
 15 #include <fstream>
 16 #include <list>
 17 #include <deque>
 18 #include <queue>
 19 #include <stack>
 20 #include <map>
 21 #include <set>
 22 #include <bitset>
 23 #include <cctype>
 24 #include <ctime>
 25 #include <utility>
 26 
 27 using namespace std;
 28 
 29 #define clr0(x) memset(x, 0, sizeof(x))
 30 #define clr1(x) memset(x, -1, sizeof(x))
 31 #define pb push_back
 32 #define sz(v) ((int)(v).size())
 33 #define all(t) t.begin(),t.end()
 34 #define INF 999999999999999999
 35 #define zero(x) (((x)>0?(x):-(x))<eps)
 36 #define out(x) cout<<#x<<":"<<(x)<<endl
 37 #define tst(a) cout<<a<<" "
 38 #define tst1(a) cout<<#a<<endl
 39 #define CINBEQUICKER std::ios::sync_with_stdio(false)
 40 
 41 const double eps = 1e-8;
 42 const double PI = atan(1.0)*4;
 43 const int inf = 2147483647 / 2;
 44 
 45 typedef vector<int> vi;
 46 typedef vector<string> vs;
 47 typedef vector<double> vd;
 48 typedef pair<int, int> pii;
 49 typedef long long int64;
 50 
 51 inline int Mymod (int a, int b) {int x=a%b; if(x<0) x+=b; return x;}
 52 
 53 bool del[4];
 54 string temp = "1689";
 55 map<int, string> mp;
 56 
 57 void gao(string s)
 58 {
 59     stringstream stm(s);
 60     int num; stm >> num;
 61     int yu = num % 7;
 62     if (!mp.count(yu)) mp[yu] = s;
 63 }
 64 
 65 int main()
 66 {
 67 //    freopen("a.in","r",stdin);
 68 //    freopen("a.out","w",stdout);
 69 //    std::ios::sync_with_stdio(false);
 70     mp.clear();
 71     string tt = "1689";
 72     gao(tt);
 73     while (next_permutation(tt.begin(), tt.end())) gao(tt);
 74 
 75     string s;
 76     while (cin >> s){
 77         clr0 (del);
 78         string ss; ss.clear();
 79         int n = sz(s), cnt = 0;
 80         for (int i = 0; i < n; ++ i){
 81             if (s[i] == '0'){
 82                 ++ cnt; continue;
 83             }
 84             bool ok = 0;
 85             for (int j = 0; j < 4; ++ j) if (s[i] == temp[j] && !del[j]){
 86                 del[j] = 1; ok = 1;
 87             }
 88             if (!ok) ss.pb (s[i]);
 89         }
 90 
 91         int len = sz(ss);
 92         if (!len){
 93             ss = mp[0];
 94             for (int i = 0; i < cnt; ++ i) ss.pb ('0');
 95             cout << ss << endl;
 96             continue;
 97         }
 98         for (int i = 0; i < cnt; ++ i) ss.pb ('0');
 99         len = sz(ss);
100         int flag = 0;
101         for (int i = 0; i < len; ++ i)
102             flag = (flag*10 + ss[i] - '0')  % 7;
103         flag = flag * 10000 % 7;
104         ss += mp[(7 - flag) % 7];
105         cout << ss << endl;
106     }
107     return 0;
108 }
View Code

 

D

题意：有一个0，1矩阵(最大5000*5000)，你可以无限次数地交换任意两行的位置。求交换之后，单个只含有1的矩形的面积最大，并返回这个面积值。

解法：枚举矩形的左下角是从哪一列开始，统计每一行从这一列开始连续的1有多少个记录在num数组里，然后从大到小遍历num数组，并更新面积值即可。

tag:dp, think, good

 1 /*
 2  * Author:  Plumrain
 3  * Created Time:  2013-12-26 12:49
 4  * File Name: D.cpp
 5  */
 6 #include <iostream>
 7 #include <cstdio>
 8 #include <cstring>
 9 #include <string>
10 #include <cmath>
11 #include <algorithm>
12 #include <vector>
13 #include <cstdlib>
14 #include <sstream>
15 #include <fstream>
16 #include <list>
17 #include <deque>
18 #include <queue>
19 #include <stack>
20 #include <map>
21 #include <set>
22 #include <bitset>
23 #include <cctype>
24 #include <ctime>
25 #include <utility>
26 
27 using namespace std;
28 
29 #define clr0(x) memset(x, 0, sizeof(x))
30 #define clr1(x) memset(x, -1, sizeof(x))
31 #define pb push_back
32 #define sz(v) ((int)(v).size())
33 #define all(t) t.begin(),t.end()
34 #define INF 999999999999999999
35 #define zero(x) (((x)>0?(x):-(x))<eps)
36 #define out(x) cout<<#x<<":"<<(x)<<endl
37 #define tst(a) cout<<a<<" "
38 #define tst1(a) cout<<#a<<endl
39 #define CINBEQUICKER std::ios::sync_with_stdio(false)
40 
41 const double eps = 1e-8;
42 const double PI = atan(1.0)*4;
43 const int inf = 2147483647 / 2;
44 
45 typedef vector<int> vi;
46 typedef vector<string> vs;
47 typedef vector<double> vd;
48 typedef pair<int, int> pii;
49 typedef long long int64;
50 
51 inline int Mymod (int a, int b) {int x=a%b; if(x<0) x+=b; return x;}
52 
53 int n, m;
54 int p[5005][5005], num[5005];
55 char v[5005][5005];
56 
57 int main()
58 {
59 //    freopen("a.in","r",stdin);
60     //freopen("a.out","w",stdout);
61 //    std::ios::sync_with_stdio(false);
62     while (scanf ("%d%d", &n, &m) != EOF){
63         string s;
64         for (int i = 0; i < n; ++ i)
65             scanf ("%s", v[i]);
66         for (int i = 0; i < n; ++ i){
67             int tmp = m;
68             for (int j = m-1; j >= 0; -- j){
69                 if (v[i][j] == '0') 
70                     tmp = j, p[i][j] = j;
71                 else
72                     p[i][j] = tmp;
73             }
74         }
75         int ans = 0;
76         for (int i = 0; i < m; ++ i){
77             clr0 (num);
78             for (int j = 0; j < n; ++ j) num[p[j][i] - i] ++;
79             int pos = 5000;
80             while (pos && num[pos] == 0) -- pos;
81             int cnt = 0;
82             while (pos){
83                 if (num[pos]) cnt += num[pos];
84                 ans = max(cnt*pos, ans);
85                 -- pos;
86             }
87         }
88         printf ("%d\n", ans);
89     }
90     return 0;
91 }
View Code

 

 

posted @ 2013-12-26 20:30  Plumrain  阅读(213)  评论(0)  编辑  收藏  举报