【CF1206B】Make Product Equal One

Description

Given a sequence of integers, each operation may be one element of a value of +1 or -1, the minimum number of seek operations may cause the product sequence requires each element is equal to 1

Solution

dp

Some of the product of the number equal to 1, a certain sequence described containing only 1 or -1, and -1 must be an even number of occurrences

Then we define $ f [i] [1/0] $ $ I $ denotes the number of front, a total of odd / even smallest operand is the number -1

So we consider the transfer

-1 if the number of times we let the current number becomes 1, then the transition from stage i-1 to i parity unchanged

Similarly, becomes -1 means that the parity change

Then the state transition equation is very simple

Code

#include <bits/stdc++.h>
namespace shl {
    typedef long long ll;
    int n, a[100010];
    ll f[100010][3];
    int c[100010][3];
    inline int read() {
        int ret = 0, op = 1;
        char c = getchar();
        while (!isdigit(c)) {
            if (c == '-') op = -1; 
            c = getchar();
        }
        while (isdigit(c)) {
            ret = (ret << 3) + (ret << 1) + c - '0';
            c = getchar();
        }
        return ret * op;
    }
    inline int aabs(int x) {
        return x >= 0 ? x : -x;
    }
    using std :: min;
    int main() {
        n = read();
        for (register int i = 1; i <= n; ++i) a[i] = read();
        for (register int i = 1; i <= n; ++i) {
            c[i][0] = aabs(a[i] + 1);   // change into -1
            c[i][1] = aabs(a[i] - 1);   // change into 1
            // std :: cerr << c[i][0] << " " << c[i][1] << std :: endl;
        }
        f[0][1] = 1ll << 60;
        for (register int i = 1; i <= n; ++i) {
            f[i][0] = min(f[i - 1][0] + c[i][1], f[i - 1][1] + c[i][0]);
            f[i][1] = min(f[i - 1][1] + c[i][1], f[i - 1][0] + c[i][0]);
        }
        printf("%I64d\n", f[n][0]);
        return 0;
    }
};
int main() {
    shl :: main();
    return 0;
}