ICM Technex 2017 and Codeforces Round #400 (Div. 1 + Div. 2, combined)

A

#include <cstdio>#include <cstring>#include <cmath>#include <cctype>#include <iostream>#include <algorithm>#include <string>#include <queue>#include <stack>#include <vector>#include <set>#include <map>#include <utility>using namespace std;typedef long long ll;const int qq = 1e5 + 10;int n, m;string a, b;string c, d;int main(){cin >> a >> b;cout << a << " " << b << endl;cin >> n;for(int i = 0; i < n; ++i){cin >> c >> d;if(c == a){a = d;cout << a << " " << b << endl;}else if(c == b){b = d;cout << a << " " << b << endl;}else if(d == a){a = c;cout << a << " " << b << endl;}else if(d == b){b = c;cout << a << " " << b << endl;}}return 0;}

B

題意:要求一個數和它的所有質因子的顏色不一樣,  要你輸出最少使用多少顏色, 以爲每一個數使用了哪種顏色

我們知道一個數的質因子之間都是互質的, 那麼我們得知如果所有質數都爲顏色1, 那麼其他數只要爲顏色2就可以得解

注意一下只需要用一種顏色的情況

#include <cstdio>#include <cstring>#include <cmath>#include <cctype>#include <iostream>#include <algorithm>#include <string>#include <queue>#include <stack>#include <vector>#include <set>#include <map>#include <utility>using namespace std;typedef long long ll;const int qq = 1e5 + 10;int n, m;int prime[30] = {2, 3, 5, 7, 11, 13, 17, 19, 21, 23, 29, 31};int vis[qq];bool isprime[qq];int main(){memset(isprime, true, sizeof(isprime));memset(vis, 0, sizeof(vis));isprime[1] = false;int n;scanf("%d", &n);if(n == 1){printf("1\n");printf("1\n");return 0;}if(n == 2){printf("1\n1 1\n");return 0;}n++;for(int i = 2; i <= n; ++i){if(isprime[i])for(int j = i + i; j <= n; j += i)isprime[j] = false;}printf("2\n");for(int i = 2; i <= n; ++i)if(isprime[i])printf("1 ");elseprintf("2 ");puts("");return 0;}

C

題意:要你求有多少個區間的和爲k^n此方(n >= 0)

這裏利用了一些數的特性, 說實話我覺得真的設計的很巧妙

對於某個前綴pre[i]我們處理出所有k^n + pre[i]的結果

那麼對於前綴pre[j] 我們知道 如果pre[j] = k^n + pre[i] 及 pre[j] - pre[i] = k^n

那麼區間[i, j] 即存在解

我們只需要統計計數就好了

#include <cstdio>#include <cstring>#include <cmath>#include <cctype>#include <iostream>#include <algorithm>#include <string>#include <queue>#include <stack>#include <vector>#include <set>#include <map>#include <utility>using namespace std;typedef long long ll;const int qq = 1e5 + 10;ll num[qq];ll powk[qq];ll pre[qq];map<ll, ll> mp;ll n, k;#define sf(x) scanf("%lld", &x);#define REP(i, x, n)for(int i = x; i < n; ++i)int main(){sf(n);sf(k);for(int i = 1; i <= n; ++i)sf(num[i]);powk[0] = 1;int cnt = 1;if(k == -1){++cnt;powk[1] = -1;}else if(k != 1){cnt = 1;ll t = 1;while(abs(t) < 1e15){powk[cnt++] = t * k;t = t * k;}}REP(i, 0, cnt)mp[powk[i]]++;ll ans = 0;for(int i = 1; i <= n; ++i){pre[i] = pre[i - 1] + num[i];ans += mp[pre[i]];for(int j = 0; j < cnt; ++j)mp[pre[i] + powk[j]]++;}printf("%lld", ans);return 0;}