題意: 給定長度爲n的數組,給定n個數字,a[i] <= a[i + 1], 問是否可以找到ai,aj,ak,使得這三個數字組不成三角形。
題解: 一開始看錯題意,找三角形,寫了半個小時發現要找不能作三角形的。只要判斷前兩個數字和最後一個數字便可。由於前兩個數字最小,最後一個數字最大。
代碼:c++
#include <bits/stdc++.h> using namespace std; int const N = 1e5 + 10; int a[N], n, T; int main() { cin >> T; while (T--) { cin >> n; for (int i = 1; i <= n; ++i) scanf("%d", &a[i]); if (a[1] + a[2] <= a[n]) cout << 1 << " " << 2 << " " << n << endl; else cout << -1 << endl; } return 0; }
題意: 給你一個長度爲n的0,1字符串,兩我的輪流選擇其中很多於一位的連續子序列,得分爲所選子序列中1個數的總和,你先選,問最多可以得到多少分。
題解: 把連續的1分段拿出來,排序完倒敘,奇數和累加就是得分
代碼:數組
#include <bits/stdc++.h> using namespace std; int const N = 1e5 + 10; int a[N], n, T; int main() { cin >> T; while (T--) { string s; cin >> s; vector<int> cnt; int l = 0, r = 0; while (s[r] == '0' && r < s.size()) r++; // cout << r << endl; l = r; // cout << l << " " << r << endl; while (1) { while (s[r] == '1' && r < s.size()) r++; // cout << r << endl; cnt.push_back(r - l); while (s[r] == '0' && r < s.size()) r++; l = r; if (r == s.size()) break; } int res = 0; sort(cnt.begin(), cnt.end()); reverse(cnt.begin(), cnt.end()); for (int i = 0; i < cnt.size(); ++i) { if (i % 2 == 0) { res += cnt[i]; } } cout << res << endl; } return 0; }
題意: 給你T組數據,每組給你一個長度爲n的字符串,求知足
\(\sum_{i=l}^{r}ai=r-l+1\) 的子區間的個數(i < j)
題解: 把\(\sum_{i=l}^{r}ai=r-l+1\)左右兩邊都減去r-l+1,則得\(\sum_{i=l}^{r}(ai-1)=0\),因此只須要找區間和爲0得個數。找區間和爲0得個數能夠使用前綴統計的方式,具體見代碼。
代碼:spa
#include <bits/stdc++.h> using namespace std; typedef long long LL; int const N = 1e5 + 10; LL T, n, a[N], sum[N]; string s; map<LL, LL> mp; int main() { cin >> T; while (T--) { cin >> n; cin >> s; mp.clear(); for (int i = 0; i < s.size(); ++i) a[i + 1] = s[i] - '0' - 1; for (int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + a[i]; // for (int i = 1; i <= n; ++i) cout << a[i] << endl; LL res = 0; // int l = 1; mp[0]++; // while (a[l] == 0 && l < n) res ++, l ++; for (int i = 1; i <= n; ++i) { // if (a[i] == 0) res++; res += mp[sum[i]]; mp[sum[i]]++; } cout << res << endl; } } // 1 2 0 // 0 1 -1 // 1 + // 0 0 0 // 11011 // 0 0 -1 0 0 // 1 + 1 + 1 + 1 +
題意: 有R,G,B三種顏色的棍子。如今告訴你每種棍子長度的對數和每對的長度,讓你用兩種不一樣顏色的棍子對圍成矩形,求每一個矩形加起來的最大值
題解: 把每一個棍子拿起來從大到小排序,而後dp暴力轉移
代碼:.net
#include <bits/stdc++.h> using namespace std; typedef long long LL; int const N = 210; LL f[N][N][N], r, g, b, A[N], B[N], C[N]; int main() { cin >> r >> g >> b; memset(f, 0, sizeof f); for (int i = 1; i <= r; ++i) scanf("%lld", &A[i]); for (int i = 1; i <= g; ++i) scanf("%lld", &B[i]); for (int i = 1; i <= b; ++i) scanf("%lld", &C[i]); sort(A + 1, A + r + 1); reverse(A + 1, A + r + 1); sort(B + 1, B + g + 1); reverse(B + 1, B + g + 1); sort(C + 1, C + b + 1); reverse(C + 1, C + b + 1); for (int i = 0; i <= r; ++i) { for (int j = 0; j <= g; ++j) { for (int k = 0; k <= b; ++k) { LL t1 = 0 , t2 = 0, t3 = 0; if (i && j) t1 = f[i - 1][j - 1][k] + A[i] * B[j]; if (i && k) t2 = f[i - 1][j][k - 1] + A[i] * C[k]; if (j && k) t3 = f[i][j - 1][k - 1] + B[j] * C[k]; f[i][j][k] = max(max(t1, f[i][j][k]), max(t2, t3)); } } } LL res = 0; for (int i = 0; i <= r; ++i) for (int j = 0; j <= g; ++j) for (int k = 0; k <= b; ++k) res = max(res, f[i][j][k]); cout << res << endl; return 0; }