CodeForces 221(div 2)

无trick水题。。。

 /*

  * Author:  Plumrain

  * Created Time:  2013-12-24 22:26

  * File Name: B.cpp

  */

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <string>

 #include <cmath>

 #include <algorithm>

 #include <vector>

 #include <cstdlib>

 #include <sstream>

 #include <fstream>

 #include <list>

 #include <deque>

 #include <queue>

 #include <stack>

 #include <map>

 #include <set>

 #include <bitset>

 #include <cctype>

 #include <ctime>

 #include <utility>

 using namespace std;

 #define clr0(x) memset(x, 0, sizeof(x))

 #define clr1(x) memset(x, -1, sizeof(x))

 #define pb push_back

 #define sz(v) ((int)(v).size())

 #define all(t) t.begin(),t.end()

 #define INF 999999999999999999

 #define zero(x) (((x)>0?(x):-(x))<eps)

 #define out(x) cout<<#x<<":"<<(x)<<endl

 #define tst(a) cout<<a<<" "

 #define tst1(a) cout<<#a<<endl

 #define CINBEQUICKER std::ios::sync_with_stdio(false)

 const double eps = 1e-;

 const double PI = atan(1.0)*;

 const int inf =  / ;

 typedef vector<int> vi;

 typedef vector<string> vs;

 typedef vector<double> vd;

 typedef pair<int, int> pii;

 typedef long long int64;

 inline int Mymod (int a, int b) {int x=a%b; if(x<) x+=b; return x;}

 int ru[], chu[];

 int main()

 {

 //    freopen("a.in","r",stdin);

 //    freopen("a.out","w",stdout);

 //    std::ios::sync_with_stdio(false);

     int n, m;

     while (scanf ("%d%d", &n , &m) != EOF){

         int t1, t2, w;

         clr0 (chu); clr0 (ru);

         for (int i = ; i < m; ++ i){

             scanf ("%d%d%d", &t1, &t2, &w);

             chu[--t1] += w;

             ru[--t2] += w;

         }

         int ans = ;

         for (int i = ; i < n; ++ i)

             ans += abs(chu[i] - ru[i]);

         printf ("%d\n", ans / );

     }

     return ;

 }

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。。。。。。

 /*

  * Author:  Plumrain

  * Created Time:  2013-12-24 22:26

  * File Name: B.cpp

  */

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <string>

 #include <cmath>

 #include <algorithm>

 #include <vector>

 #include <cstdlib>

 #include <sstream>

 #include <fstream>

 #include <list>

 #include <deque>

 #include <queue>

 #include <stack>

 #include <map>

 #include <set>

 #include <bitset>

 #include <cctype>

 #include <ctime>

 #include <utility>

 using namespace std;

 #define clr0(x) memset(x, 0, sizeof(x))

 #define clr1(x) memset(x, -1, sizeof(x))

 #define pb push_back

 #define sz(v) ((int)(v).size())

 #define all(t) t.begin(),t.end()

 #define INF 999999999999999999

 #define zero(x) (((x)>0?(x):-(x))<eps)

 #define out(x) cout<<#x<<":"<<(x)<<endl

 #define tst(a) cout<<a<<" "

 #define tst1(a) cout<<#a<<endl

 #define CINBEQUICKER std::ios::sync_with_stdio(false)

 const double eps = 1e-;

 const double PI = atan(1.0)*;

 const int inf =  / ;

 typedef vector<int> vi;

 typedef vector<string> vs;

 typedef vector<double> vd;

 typedef pair<int, int> pii;

 typedef long long int64;

 inline int Mymod (int a, int b) {int x=a%b; if(x<) x+=b; return x;}

 int ru[], chu[];

 int main()

 {

 //    freopen("a.in","r",stdin);

 //    freopen("a.out","w",stdout);

 //    std::ios::sync_with_stdio(false);

     int n, m;

     while (scanf ("%d%d", &n , &m) != EOF){

         int t1, t2, w;

         clr0 (chu); clr0 (ru);

         for (int i = ; i < m; ++ i){

             scanf ("%d%d%d", &t1, &t2, &w);

             chu[--t1] += w;

             ru[--t2] += w;

         }

         int ans = ;

         for (int i = ; i < n; ++ i)

             ans += abs(chu[i] - ru[i]);

         printf ("%d\n", ans / );

     }

     return ;

 }

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

 /*

  * Author:  Plumrain

  * Created Time:  2013-12-25 14:09

  * File Name: C.cpp

  */

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <string>

 #include <cmath>

 #include <algorithm>

 #include <vector>

 #include <cstdlib>

 #include <sstream>

 #include <fstream>

 #include <list>

 #include <deque>

 #include <queue>

 #include <stack>

 #include <map>

 #include <set>

 #include <bitset>

 #include <cctype>

 #include <ctime>

 #include <utility>

 using namespace std;

 #define clr0(x) memset(x, 0, sizeof(x))

 #define clr1(x) memset(x, -1, sizeof(x))

 #define pb push_back

 #define sz(v) ((int)(v).size())

 #define all(t) t.begin(),t.end()

 #define INF 999999999999999999

 #define zero(x) (((x)>0?(x):-(x))<eps)

 #define out(x) cout<<#x<<":"<<(x)<<endl

 #define tst(a) cout<<a<<" "

 #define tst1(a) cout<<#a<<endl

 #define CINBEQUICKER std::ios::sync_with_stdio(false)

 const double eps = 1e-;

 const double PI = atan(1.0)*;

 const int inf =  / ;

 typedef vector<int> vi;

 typedef vector<string> vs;

 typedef vector<double> vd;

 typedef pair<int, int> pii;

 typedef long long int64;

 inline int Mymod (int a, int b) {int x=a%b; if(x<) x+=b; return x;}

 bool del[];

 string temp = "";

 map<int, string> mp;

 void gao(string s)

 {

     stringstream stm(s);

     int num; stm >> num;

     int yu = num % ;

     if (!mp.count(yu)) mp[yu] = s;

 }

 int main()

 {

 //    freopen("a.in","r",stdin);

 //    freopen("a.out","w",stdout);

 //    std::ios::sync_with_stdio(false);

     mp.clear();

     string tt = "";

     gao(tt);

     while (next_permutation(tt.begin(), tt.end())) gao(tt);

     string s;

     while (cin >> s){

         clr0 (del);

         string ss; ss.clear();

         int n = sz(s), cnt = ;

         for (int i = ; i < n; ++ i){

             if (s[i] == ''){

                 ++ cnt; continue;

             }

             bool ok = ;

             for (int j = ; j < ; ++ j) if (s[i] == temp[j] && !del[j]){

                 del[j] = ; ok = ;

             }

             if (!ok) ss.pb (s[i]);

         }

         int len = sz(ss);

         if (!len){

             ss = mp[];

             for (int i = ; i < cnt; ++ i) ss.pb ('');

             cout << ss << endl;

             continue;

         }

         for (int i = ; i < cnt; ++ i) ss.pb ('');

         len = sz(ss);

         int flag = ;

         for (int i = ; i < len; ++ i)

             flag = (flag* + ss[i] - '')  % ;

         flag = flag *  % ;

         ss += mp[( - flag) % ];

         cout << ss << endl;

     }

     return ;

 }

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

 /*

  * Author:  Plumrain

  * Created Time:  2013-12-26 12:49

  * File Name: D.cpp

  */

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <string>

 #include <cmath>

 #include <algorithm>

 #include <vector>

 #include <cstdlib>

 #include <sstream>

 #include <fstream>

 #include <list>

 #include <deque>

 #include <queue>

 #include <stack>

 #include <map>

 #include <set>

 #include <bitset>

 #include <cctype>

 #include <ctime>

 #include <utility>

 using namespace std;

 #define clr0(x) memset(x, 0, sizeof(x))

 #define clr1(x) memset(x, -1, sizeof(x))

 #define pb push_back

 #define sz(v) ((int)(v).size())

 #define all(t) t.begin(),t.end()

 #define INF 999999999999999999

 #define zero(x) (((x)>0?(x):-(x))<eps)

 #define out(x) cout<<#x<<":"<<(x)<<endl

 #define tst(a) cout<<a<<" "

 #define tst1(a) cout<<#a<<endl

 #define CINBEQUICKER std::ios::sync_with_stdio(false)

 const double eps = 1e-;

 const double PI = atan(1.0)*;

 const int inf =  / ;

 typedef vector<int> vi;

 typedef vector<string> vs;

 typedef vector<double> vd;

 typedef pair<int, int> pii;

 typedef long long int64;

 inline int Mymod (int a, int b) {int x=a%b; if(x<) x+=b; return x;}

 int n, m;

 int p[][], num[];

 char v[][];

 int main()

 {

 //    freopen("a.in","r",stdin);

     //freopen("a.out","w",stdout);

 //    std::ios::sync_with_stdio(false);

     while (scanf ("%d%d", &n, &m) != EOF){

         string s;

         for (int i = ; i < n; ++ i)

             scanf ("%s", v[i]);

         for (int i = ; i < n; ++ i){

             int tmp = m;

             for (int j = m-; j >= ; -- j){

                 if (v[i][j] == '')

                     tmp = j, p[i][j] = j;

                 else

                     p[i][j] = tmp;

             }

         }

         int ans = ;

         for (int i = ; i < m; ++ i){

             clr0 (num);

             for (int j = ; j < n; ++ j) num[p[j][i] - i] ++;

             int pos = ;

             while (pos && num[pos] == ) -- pos;

             int cnt = ;

             while (pos){

                 if (num[pos]) cnt += num[pos];

                 ans = max(cnt*pos, ans);

                 -- pos;

             }

         }

         printf ("%d\n", ans);

     }

     return ;

 }

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