Educational Codeforces Round 21 Problem A - C

Problem A Lucky Year

题目传送门[here]

　　题目大意是说，只有一个数字非零的数是幸运的，给出一个数，求下一个幸运的数是多少。

　　这个幸运的数不是最高位的数字都是零，于是只跟最高位有关，只保留最高位，然后加一求差就是答案。

Code

 /**
  * Codeforces
  * Problem#808A
  * Accepted
  * Time:15ms
  * Memory:0k
  */
 #include<iostream>
 #include<cstdio>
 #include<ctime>
 #include<cctype>
 #include<cstring>
 #include<cstdlib>
 #include<fstream>
 #include<sstream>
 #include<algorithm>
 #include<map>
 #include<set>
 #include<stack>
 #include<queue>
 #include<vector>
 #include<stack>
 using namespace std;
 typedef bool boolean;
 #define inf 0xfffffff
 #define smin(a, b) a = min(a, b)
 #define smax(a, b) a = max(a, b)
 #define max3(a, b, c) max(a, max(b, c))
 #define min3(a, b, c) min(a, min(b, c))
 template<typename T>
 inline boolean readInteger(T& u){
     char x;
     int aFlag = ;
     while(!isdigit((x = getchar())) && x != '-' && x != -);
     if(x == -) {
         ungetc(x, stdin);
         return false;
     }
     if(x == '-'){
         x = getchar();
         aFlag = -;
     }
     for(u = x - ''; isdigit((x = getchar())); u = (u << ) + (u << ) + x - '');
     ungetc(x, stdin);
     u *= aFlag;
     return true;
 }

 int power[];

 int n;
 int bits = ;

 inline void init() {
     power[] = ;
     for(int i = ; i <= ; i++)
         power[i] = power[i - ] * ;
 }

 inline void solve() {
     readInteger(n);
     while(power[bits] <= n)    bits++;
     int high = n - n % power[bits - ];
     int next = high + power[bits - ];
     printf("%d\n", next - n);
 }

 int main() {
     init();
     solve();
     return ;
 }

Problem A

---

Problem B Average Sleep Time

题目传送门[here]

　　题目大意是说，给出n个数a_i和k，再得到n - k + 1个新数，第i个新数为,再求这几个新数的平均数。

　　有两种方法，第一种是前缀和暴力乱搞。第二种是特殊处理两段的数被求和的次数，中间的都是k，然后就可以求和了，然后就可以求平均数。

　　我呢，用的第一种方法，因为懒。

Code

 /**
  * Codeforces
  * Problem#808B
  * Accepted
  * Time:30ms
  * Memory:2400k
  */
 #include<iostream>
 #include<cstdio>
 #include<ctime>
 #include<cctype>
 #include<cstring>
 #include<cstdlib>
 #include<fstream>
 #include<sstream>
 #include<algorithm>
 #include<map>
 #include<set>
 #include<stack>
 #include<queue>
 #include<vector>
 #include<stack>
 using namespace std;
 typedef bool boolean;
 #define inf 0xfffffff
 #define smin(a, b) a = min(a, b)
 #define smax(a, b) a = max(a, b)
 #define max3(a, b, c) max(a, max(b, c))
 #define min3(a, b, c) min(a, min(b, c))
 template<typename T>
 inline boolean readInteger(T& u){
     char x;
     int aFlag = ;
     while(!isdigit((x = getchar())) && x != '-' && x != -);
     if(x == -) {
         ungetc(x, stdin);
         return false;
     }
     if(x == '-'){
         x = getchar();
         aFlag = -;
     }
     for(u = x - ''; isdigit((x = getchar())); u = (u << ) + (u << ) + x - '');
     ungetc(x, stdin);
     u *= aFlag;
     return true;
 }

 #define LL long long

 int n, k;
 LL *sum;
 int* a;
 double w;

 inline void init() {
     readInteger(n);
     readInteger(k);
     w = n - k + ;
     sum = new LL[(const int)(n + )];
     a = new int[(const int)(n + )];
     sum[] = ;
     for(int i = ; i <= n; i++) {
         readInteger(a[i]);
         sum[i] = sum[i - ] + a[i];
     }
 }

 LL s = ;
 double avg = 0.0;

 inline void solve() {
     for(int i = k; i <= n; i++) {
         s += sum[i] - sum[i - k];
     }
     avg = s / w;
     printf("%.9lf", avg);
 }

 int main() {
     init();
     solve();
     return ;
 }

Problem B

---

Problem C Tea Party

题目传送门[here]

　　语文不好，题目大意就不给了(显然是太懒了)，去看原题吧，看不懂扔给谷歌机翻。

　　首先呢给每人达到最低要求的茶叶量，这时判一下是否合法。(然后跳过一个if吧)。

　　接着很容易想到一个符合题意的贪心，给茶杯最大的人加尽可能多的茶叶(能加满就加满)，如果还有剩的，就给茶杯第二大的人加....

　　这个显然是合法，不会出现让顾客不满意的情况。

Code

 /**
  * Codeforces
  * Problem#808C
  * Accepted
  * Time:15ms
  * Memory:0k
  */
 #include<iostream>
 #include<cstdio>
 #include<ctime>
 #include<cctype>
 #include<cstring>
 #include<cstdlib>
 #include<fstream>
 #include<sstream>
 #include<algorithm>
 #include<map>
 #include<set>
 #include<stack>
 #include<queue>
 #include<vector>
 #include<stack>
 using namespace std;
 typedef bool boolean;
 #define inf 0xfffffff
 #define smin(a, b) a = min(a, b)
 #define smax(a, b) a = max(a, b)
 #define max3(a, b, c) max(a, max(b, c))
 #define min3(a, b, c) min(a, min(b, c))
 template<typename T>
 inline boolean readInteger(T& u){
     char x;
     int aFlag = ;
     while(!isdigit((x = getchar())) && x != '-' && x != -);
     if(x == -) {
         ungetc(x, stdin);
         return false;
     }
     if(x == '-'){
         x = getchar();
         aFlag = -;
     }
     for(u = x - ''; isdigit((x = getchar())); u = (u << ) + (u << ) + x - '');
     ungetc(x, stdin);
     u *= aFlag;
     return true;
 }

 typedef class Teacup {
     public:
         int val;
         int pos;

         Teacup(int val = , int pos = ):val(val), pos(pos) {        }

         boolean operator < (Teacup b) const {
             if(val != b.val)    return val > b.val;
             return pos < b.pos;
         }
 }Teacup;

 int n;
 int w;
 int *a;
 int s = ;
 int *b;
 Teacup* tc;

 inline void init() {
     readInteger(n);
     readInteger(w);
     a = new int[(const int)(n + )];
     b = new int[(const int)(n + )];
     for(int i = ; i <= n; i++) {
         readInteger(a[i]);
         b[i] = (a[i] + ) / ;
         s += b[i];
     }
 }

 inline void solve() {
     if(s > w) {
         puts("-1");
         return;
     }
     w -= s;
     tc = new Teacup[(const int)(n + )];
     for(int i = ; i <= n; i++)
         tc[i] = Teacup(a[i], i);
     sort(tc + , tc + n + );
     int i = ;
     while(w) {
         i++;
         int delta = min(w, tc[i].val - b[tc[i].pos]);
         b[tc[i].pos] += delta;
         w -= delta;
     }
     for(int i = ; i <= n; i++) {
         printf("%d ", b[i]);
     }
 }

 int main() {
     init();
     solve();
     return ;
 }

Problem C