Vasya has n pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. When he comes home in the evening, Vasya takes off the used socks and throws them away. Every m-th day (at days with numbers m,?2m,?3m,?...) mom buys a pair of socks to Vasya. She does it late in the evening, so that Vasya cannot put on a new pair of socks before the next day. How many consecutive days pass until Vasya runs out of socks?
InputThe single line contains two integers n and m (1?≤?n?≤?100; 2?≤?m?≤?100), separated by a space.
OutputPrint a single integer — the answer to the problem.
Sample test(s) input2 2output
3input
9 3output
13傳送門:點擊打開鏈接 解題思路: 水題,簡單模擬。 代碼:
#include#include #include #include #include using namespace std; typedef long long lint; typedef double DB; //const int MAXN = ; int main() { int n, m, t = 0; scanf("%d%d", &n, &m); while(n) { n--; t++; if(0 == t%m) n++; } printf("%d\n", t); return 0; }
Little Dima misbehaved during a math lesson a lot and the nasty teacher Mr. Pickles gave him the following problem as a punishment.
Find all integer solutions x (0?x?109) of the equation:
where a, b, c are some predetermined constant values and function s(x) determines the sum of all digits in the decimal representation of number x.
The teacher gives this problem to Dima for each lesson. He changes only the parameters of the equation: a, b, c. Dima got sick of getting bad marks and he asks you to help him solve this challenging problem.
InputThe first line contains three space-separated integers: a,?b,?c (1?≤?a?≤?5; 1?≤?b?≤?10000; ?-?10000?≤?c?≤?10000).
OutputPrint integer n — the number of the solutions that you've found. Next print n integers in the increasing order — the solutions of the given equation. Print only integer solutions that are larger than zero and strictly less than 109.
Sample test(s) input3 2 8output
3 10 2008 13726input
1 2 -18output
0input
2 2 -1output
4 1 31 337 967
#include#include #include #include #include using namespace std; typedef long long lint; typedef double DB; const int MAX = 1e9; const int MAXN = 100; lint ans[100]; int fun(lint x) { int ret = 0; while(x) { ret += x%10; x /= 10; } return ret; } int main() { int a, b, c, n, m = 0; scanf("%d%d%d", &a, &b, &c); for(int i=1; i<=81; ++i) { lint x = 1ll*b*pow(i*1.0,a) + 1ll*c; if(x 0 && i==fun(x)) ans[m++] = x; } sort(ans, ans+m); printf("%d\n", m); for(int i=0; i
C. Present time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard outputLittle beaver is a beginner programmer, so informatics is his favorite subject. Soon his informatics teacher is going to have a birthday and the beaver has decided to prepare a present for her. He planted n flowers in a row on his windowsill and started waiting for them to grow. However, after some time the beaver noticed that the flowers stopped growing. The beaver thinks it is bad manners to present little flowers. So he decided to come up with some solutions.
There are m days left to the birthday. The height of the i-th flower (assume that the flowers in the row are numbered from 1 to n from left to right) is equal to ai at the moment. At each of the remaining m days the beaver can take a special watering and water w contiguous flowers (he can do that only once at a day). At that each watered flower grows by one height unit on that day. The beaver wants the height of the smallest flower be as large as possible in the end. What maximum height of the smallest flower can he get?
InputThe first line contains space-separated integers n, m and w (1?≤?w?≤?n?≤?105; 1?≤?m?≤?105). The second line contains space-separated integers a1,?a2,?...,?an (1?≤?ai?≤?109).
OutputPrint a single integer — the maximum final height of the smallest flower.
Sample test(s) input6 2 3 22 2 2 1 1output2input2 5 1 5 8output9傳送門:點擊打開鏈接 解題思路: 對所求解的值進行二分。ps:這裡的b數組大小是n+w。 代碼:#include#include #include #include #include using namespace std; typedef long long lint; typedef double DB; const int MAXN = 2e5+10; const int INF = 2e9; lint a[MAXN], b[MAXN], ans; int n, m, w; bool check(lint k) { memset(b, 0, sizeof(b)); lint sum = 0, d = 0; for(int i=1; i<=n; ++i) { sum += b[i]; lint tp = k - a[i] - sum; if(tp > 0) { sum += tp; b[i+w] -= tp; d += tp; // printf("%I64d %I64d\n", tp, d); if(d > m) return false; } } return true; } int main() { scanf("%d%d%d", &n, &m, &w); for(int i=1; i<=n; ++i) scanf("%I64d", a+i); lint l = 1, r = 1ll*INF; while(l <= r) { lint mid = (l+r)>>1; if(check(mid)) l = mid + 1, ans = mid; else r = mid - 1; // printf("%I64d %I64d\n", l, r); } printf("%I64d\n", ans); return 0; }