南京邀请赛重现

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只做了四道水题

ans = (m/n)ans + avg

注意m == n 和 avg == 0.0的情况

#include <set>#include <cmath>#include <queue>#include <stack>#include <string>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef  long long LL;const double PI = acos(-1.0);template <class T> inline  T MAX(T a, T b){if (a > b) return a;return b;}template <class T> inline  T MIN(T a, T b){if (a < b) return a;return b;}const int N = 111;const int M = 11111;const LL MOD = 1000000007LL;const int dir[4][2] = {1, 0, -1, 0, 0, -1, 0, 1};const int INF = 0x3f3f3f3f;int a[222];int main(){    int n, m;    while (scanf("%d", &n) != EOF)    {        int i, j , k, sum = 0;        double ans;        for (i = 0; i < n; ++i)        {            scanf("%d", &a[i]);            sum += a[i];        }        scanf("%d", &m);        for (i = 0; i < m; ++i)            scanf("%d", &k);        if (sum == 0)        {            printf("0.00\n");        }        else if (n == m)        {            printf("inf\n");        }        else        {            ans = sum;            ans = ans / (n - m);            printf("%.2lf\n", ans);        }    }    return 0;}

C 统计进位，只要预处理1-100000000的前缀和中每个位有多少个一，然后每次询问b-(a-1)再统计即可

#include <set>#include <cmath>#include <queue>#include <stack>#include <string>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef  long long LL;const double PI = acos(-1.0);template <class T> inline  T MAX(T a, T b){if (a > b) return a;return b;}template <class T> inline  T MIN(T a, T b){if (a < b) return a;return b;}const int N = 111;const int M = 11111;const LL MOD = 1000000007LL;const int dir[4][2] = {1, 0, -1, 0, 0, -1, 0, 1};const int INF = 0x3f3f3f3f;void solve(LL cnt[], LL num){    LL i, j, k, bas = 1, lim = 0;    for (i = 0; i <= 64; ++i)    {        if (num <= lim) break;        cnt[i] = (num - lim) / (bas * 2) * bas;        if ((num - lim) % (bas * 2) >= bas) cnt[i] = cnt[i] + bas;        else cnt[i] = cnt[i] + ((num-lim) %(bas * 2));        bas *= 2; lim = bas - 1;    }}LL c1[100], c2[100];int main(){    LL a, b;    while (scanf("%I64d%I64d", &a, &b) != EOF)    {        memset(c1, 0, sizeof(c1));memset(c2, 0, sizeof(c2));        solve(c1, a - 1); solve(c2, b);        LL tot = 0, ans = 0;//        for (int i = 0; i <= 10; ++i)//            printf("%d ", c1[i]);//        printf("\n");//        for (int i = 0; i <= 10; ++i)//            printf("%d ", c2[i]);//        printf("\n");        for (int i = 0; i <= 64; ++i)            c2[i] = c2[i] - c1[i];        for (int i = 0; i <= 64; ++i)        {            ans = ans + c2[i] / 2;            c2[i + 1] += c2[i] / 2;        }        printf("%I64d\n", ans);    }    return 0;}

H 统计哪个数字重复

#include <set>#include <cmath>#include <queue>#include <stack>#include <string>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef  long long LL;const double PI = acos(-1.0);template <class T> inline  T MAX(T a, T b){if (a > b) return a;return b;}template <class T> inline  T MIN(T a, T b){if (a < b) return a;return b;}const int N = 111;const int M = 11111;const LL MOD = 1000000007LL;const int dir[4][2] = {1, 0, -1, 0, 0, -1, 0, 1};const int INF = 0x3f3f3f3f;int a[1111];int main(){    int n;    while (scanf("%d", &n) != EOF)    {        int i, x, t;        memset(a, 0, sizeof(a));        for (i = 0; i <= n; ++i)        {            scanf("%d", &x);            if (a[x]) t = x;            a[x]++;        }        printf("%d\n", t);    }    return 0;}

枚举两个区间的X的gcd的倍数在不在两个区间只差的区间内，【Z[j]-Y[i]，Y[j] - Z[i]】

#include <set>#include <cmath>#include <queue>#include <stack>#include <string>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef  long long LL;const double PI = acos(-1.0);template <class T> inline  T MAX(T a, T b){if (a > b) return a;return b;}template <class T> inline  T MIN(T a, T b){if (a < b) return a;return b;}const int N = 111;const int M = 11111;const LL MOD = 1000000007LL;const int dir[4][2] = {1, 0, -1, 0, 0, -1, 0, 1};const int INF = 0x3f3f3f3f;int X[1111], Y[1111], Z[1111];inline int gcd(int a, int b){    if (!b) return a;    return gcd(b, a % b);}bool check(int i, int j){    if (Z[i] < Y[j] || Y[i] > Z[j])    {        int k;        int d = gcd(X[i], X[j]);        if (Z[i] > Z[j]) {k = i; i = j; j = k;}        int l, r;        l = Y[j] - Z[i]; r = Z[j] - Y[i];//        printf("%d %d %d\n", d, l, r);        if (l % d == 0 || r % d == 0)            return true;        else            return l / d != r / d;    }    return true;}int main(){    int n;    while (scanf("%d", &n) != EOF)    {        int i, j, k;        for (i = 0; i < n; ++i)            scanf("%d%d%d", &X[i], &Y[i], &Z[i]);        bool flag = true;        for (i = 0; i < n; ++i)            for (j = i + 1; j < n; ++j)            {                if (check(i, j))                {                    flag = false;                    break;                }            }        if (flag)  printf("Can Take off\n");        else printf("Cannot Take off\n");    }    return 0;}