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
1 /** 2 * Codeforces 3 * Problem#808A 4 * Accepted 5 * Time:15ms 6 * Memory:0k 7 */ 8 #include<iostream> 9 #include<cstdio> 10 #include<ctime> 11 #include<cctype> 12 #include<cstring> 13 #include<cstdlib> 14 #include<fstream> 15 #include<sstream> 16 #include<algorithm> 17 #include<map> 18 #include<set> 19 #include<stack> 20 #include<queue> 21 #include<vector> 22 #include<stack> 23 using namespace std; 24 typedef bool boolean; 25 #define inf 0xfffffff 26 #define smin(a, b) a = min(a, b) 27 #define smax(a, b) a = max(a, b) 28 #define max3(a, b, c) max(a, max(b, c)) 29 #define min3(a, b, c) min(a, min(b, c)) 30 template<typename T> 31 inline boolean readInteger(T& u){ 32 char x; 33 int aFlag = 1; 34 while(!isdigit((x = getchar())) && x != '-' && x != -1); 35 if(x == -1) { 36 ungetc(x, stdin); 37 return false; 38 } 39 if(x == '-'){ 40 x = getchar(); 41 aFlag = -1; 42 } 43 for(u = x - '0'; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - '0'); 44 ungetc(x, stdin); 45 u *= aFlag; 46 return true; 47 } 48 49 int power[10]; 50 51 int n; 52 int bits = 0; 53 54 inline void init() { 55 power[0] = 1; 56 for(int i = 1; i <= 10; i++) 57 power[i] = power[i - 1] * 10; 58 } 59 60 inline void solve() { 61 readInteger(n); 62 while(power[bits] <= n) bits++; 63 int high = n - n % power[bits - 1]; 64 int next = high + power[bits - 1]; 65 printf("%d ", next - n); 66 } 67 68 int main() { 69 init(); 70 solve(); 71 return 0; 72 }
Problem B Average Sleep Time
题目传送门[here]
题目大意是说，给出n个数ai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
1 /** 2 * Codeforces 3 * Problem#808B 4 * Accepted 5 * Time:30ms 6 * Memory:2400k 7 */ 8 #include<iostream> 9 #include<cstdio> 10 #include<ctime> 11 #include<cctype> 12 #include<cstring> 13 #include<cstdlib> 14 #include<fstream> 15 #include<sstream> 16 #include<algorithm> 17 #include<map> 18 #include<set> 19 #include<stack> 20 #include<queue> 21 #include<vector> 22 #include<stack> 23 using namespace std; 24 typedef bool boolean; 25 #define inf 0xfffffff 26 #define smin(a, b) a = min(a, b) 27 #define smax(a, b) a = max(a, b) 28 #define max3(a, b, c) max(a, max(b, c)) 29 #define min3(a, b, c) min(a, min(b, c)) 30 template<typename T> 31 inline boolean readInteger(T& u){ 32 char x; 33 int aFlag = 1; 34 while(!isdigit((x = getchar())) && x != '-' && x != -1); 35 if(x == -1) { 36 ungetc(x, stdin); 37 return false; 38 } 39 if(x == '-'){ 40 x = getchar(); 41 aFlag = -1; 42 } 43 for(u = x - '0'; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - '0'); 44 ungetc(x, stdin); 45 u *= aFlag; 46 return true; 47 } 48 49 #define LL long long 50 51 int n, k; 52 LL *sum; 53 int* a; 54 double w; 55 56 inline void init() { 57 readInteger(n); 58 readInteger(k); 59 w = n - k + 1; 60 sum = new LL[(const int)(n + 1)]; 61 a = new int[(const int)(n + 1)]; 62 sum[0] = 0; 63 for(int i = 1; i <= n; i++) { 64 readInteger(a[i]); 65 sum[i] = sum[i - 1] + a[i]; 66 } 67 } 68 69 LL s = 0; 70 double avg = 0.0; 71 72 inline void solve() { 73 for(int i = k; i <= n; i++) { 74 s += sum[i] - sum[i - k]; 75 } 76 avg = s / w; 77 printf("%.9lf", avg); 78 } 79 80 int main() { 81 init(); 82 solve(); 83 return 0; 84 }
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
1 /** 2 * Codeforces 3 * Problem#808C 4 * Accepted 5 * Time:15ms 6 * Memory:0k 7 */ 8 #include<iostream> 9 #include<cstdio> 10 #include<ctime> 11 #include<cctype> 12 #include<cstring> 13 #include<cstdlib> 14 #include<fstream> 15 #include<sstream> 16 #include<algorithm> 17 #include<map> 18 #include<set> 19 #include<stack> 20 #include<queue> 21 #include<vector> 22 #include<stack> 23 using namespace std; 24 typedef bool boolean; 25 #define inf 0xfffffff 26 #define smin(a, b) a = min(a, b) 27 #define smax(a, b) a = max(a, b) 28 #define max3(a, b, c) max(a, max(b, c)) 29 #define min3(a, b, c) min(a, min(b, c)) 30 template<typename T> 31 inline boolean readInteger(T& u){ 32 char x; 33 int aFlag = 1; 34 while(!isdigit((x = getchar())) && x != '-' && x != -1); 35 if(x == -1) { 36 ungetc(x, stdin); 37 return false; 38 } 39 if(x == '-'){ 40 x = getchar(); 41 aFlag = -1; 42 } 43 for(u = x - '0'; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - '0'); 44 ungetc(x, stdin); 45 u *= aFlag; 46 return true; 47 } 48 49 typedef class Teacup { 50 public: 51 int val; 52 int pos; 53 54 Teacup(int val = 0, int pos = 0):val(val), pos(pos) { } 55 56 boolean operator < (Teacup b) const { 57 if(val != b.val) return val > b.val; 58 return pos < b.pos; 59 } 60 }Teacup; 61 62 int n; 63 int w; 64 int *a; 65 int s = 0; 66 int *b; 67 Teacup* tc; 68 69 inline void init() { 70 readInteger(n); 71 readInteger(w); 72 a = new int[(const int)(n + 1)]; 73 b = new int[(const int)(n + 1)]; 74 for(int i = 1; i <= n; i++) { 75 readInteger(a[i]); 76 b[i] = (a[i] + 1) / 2; 77 s += b[i]; 78 } 79 } 80 81 inline void solve() { 82 if(s > w) { 83 puts("-1"); 84 return; 85 } 86 w -= s; 87 tc = new Teacup[(const int)(n + 1)]; 88 for(int i = 1; i <= n; i++) 89 tc[i] = Teacup(a[i], i); 90 sort(tc + 1, tc + n + 1); 91 int i = 0; 92 while(w) { 93 i++; 94 int delta = min(w, tc[i].val - b[tc[i].pos]); 95 b[tc[i].pos] += delta; 96 w -= delta; 97 } 98 for(int i = 1; i <= n; i++) { 99 printf("%d ", b[i]); 100 } 101 } 102 103 int main() { 104 init(); 105 solve(); 106 return 0; 107 }