CF888E Maximum Subsequence (折半枚舉+ two-pointers)
阿新 • • 發佈:2018-10-16
練手題 ces amp getchar() code two-point print clas max
題意
給定一個包含\(n\)個數的序列\(a\),在其中任選若幹個數,使得他們的和對\(m\)取模後最大。(\(n\leq 35\))
題解
顯然,\(2^n\)的暴枚是不現實的...,於是我們想到了折半枚舉,分成兩部分暴枚,然後考慮合並,合並的時候用two-pointers思想掃一遍就行了。
其實這是一道折半枚舉+Two-Pointers的很好的練手題
//最近CodeForces有點萎,可能會JudgementError,但已經評測過了,能AC,多交幾次應該可以 #include <cstdio> #include <algorithm> using std::max; using std::sort; const int N = 40, K = 19; int n, m, k, ans, totx, toty; long long a[N], x[1 << K], y[1 << K]; template <typename T> inline void read(T &x) { x = 0; char ch = getchar(); while (ch < '0' || ch > '9') ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); } void dfsx (int i, long long sum) { if (i > k) { x[++totx] = sum % m; return ; } dfsx (i + 1, sum + a[i]); dfsx (i + 1, sum); } void dfsy (int i, long long sum) { if (i > n) { y[++toty] = sum % m; return ; } dfsy (i + 1, sum + a[i]); dfsy (i + 1, sum); } int main () { read(n), read(m), k = n >> 1; for (int i = 1; i <= n; ++i) read(a[i]); dfsx(1, 0), dfsy(k + 1, 0); sort(&x[1], &x[totx + 1]), sort(&y[1], &y[toty + 1]); int l = 1, r = toty; while (l <= totx) { while (r && x[l] + y[r] >= m) --r; if(!r) break; ans = max(ans, int((x[l] + y[r]) % m)), ++l; } printf ("%d\n", ans); return 0; }
CF888E Maximum Subsequence (折半枚舉+ two-pointers)