1. 程式人生 > >CF888E Maximum Subsequence (折半枚舉+ two-pointers)

CF888E Maximum Subsequence (折半枚舉+ two-pointers)

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)