只做了四道水題

A

ans = (m/n)ans + avg

注意m == n 和 avg == 0.0的情況

#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 統計進位，只要預處理1-100000000的字首和中每個位有多少個一，然後每次詢問b-(a-1)再統計即可

#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 統計哪個數字重複

#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;
}
 

K

列舉兩個區間的X的gcd的倍數在不在 兩個區間只差的區間內，【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;
}