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  • Codeforces Round #585 (Div. 2)

    B. The Number of Products 递推:ne[i]表示以ai结尾的负区间个数,po[i]表示以ai结尾的负区间个数 ``` /**/ #include #include #include #include #include #include #include #include #include #include #include #include

    typedef long long LL;
    typedef unsigned long long ULL;
    using namespace std;

    bool Sqrt(LL n) { return (LL)sqrt(n) * sqrt(n) == n; }
    const double PI = acos(-1.0), ESP = 1e-10;
    const LL INF = 99999999999999;
    const int inf = 999999999, N = 2e5 + 24;
    int n;
    LL a[N], po[N], ne[N];

    int main()
    {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    while(scanf("%d", &n) == 1) {
    for(int i = 0; i < n; i++) scanf("%lld", &a[i]);
    memset(po, 0, sizeof po);
    memset(ne, 0, sizeof ne);
    if(a[0] < 0) ne[0] = 1;
    else po[0] = 1;
    for(int i = 1; i < n; i++) {
    if(a[i] < 0) {
    po[i] = ne[i - 1];
    ne[i] = po[i - 1] + 1;
    }
    else {
    po[i] = po[i - 1] + 1;
    ne[i] = ne[i - 1];
    }
    }
    for(int i = 0; i < n; i++) {
    po[n] += po[i];
    ne[n] += ne[i];
    }
    printf("%lld %lld ", ne[n], po[n]);
    }

    return 0;
    

    }
    /*
    input:
    output:
    modeling:
    methods:
    complexity:
    summary:
    */

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  • 原文地址:https://www.cnblogs.com/000what/p/12229358.html
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