題目連結：A題

題目大意：問有多少個小於2n的形式是2k−1的能整除7的數字的個數

題目思路：7的二進位制是111，而2k−1的二進位制是k個1，所以只需要k能被3整除就好了

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <cstring>
#include <cstdlib>
 

#include <iostream>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 1e6+10;

int T,n,m;

int main(){
    scanf("%d",&T);
    for(int Case = 1;Case <= T;Case++){
        scanf("%d",&n);
        printf("Case #%d: %d\n",Case,n/3);
    }
    return
 
 0;
}

題目連結：D題

題目大意：有n個蛋糕，每個蛋糕都有尺寸，現在問你能形成多少個蛋糕塔，要求：蛋糕塔的層數一定得是k，每一層只有能一個蛋糕，下一層的蛋糕得尺寸至少得是上一層得兩倍

題目思路：二分蛋糕塔的個數，然後我們優先取最小的mid個，然後線性掃看能合成多少層，然後層數和k比較得二分check性

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
 

#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

using namespace std;
typedef long long ll;
const ll maxn = 1e6+10;

ll T,a[maxn],b[maxn],n,k;

bool check(ll mid){
    if(mid == 0) return true;
    memset(b,0,sizeof(b));
    ll cnt = 0,now = 0;
    for(ll i = 0;i < n;i++){
        if(a[i] >= b[now]*2){
            b[now] = a[i];
            now++;
            if(now == mid){
                cnt++;
                now = 0;
            }
        }
    }
    if(cnt >= k) return true;
    return false;
}

int main(){
    scanf("%lld",&T);
    for(ll Case = 1;Case <= T;Case++){
        scanf("%lld%lld",&n,&k);
        for(ll i = 0;i < n;i++) scanf("%lld",&a[i]);
        sort(a,a+n);
        ll l = 0,r = 1e6+10;
        ll mid = (l+r)/2,maxx = -1;
        while(l <= r){
            mid = (l+r)/2;
            if(check(mid)){
                maxx = max(maxx,mid);
                l = mid+1;
            }
            else r = mid-1;
        }
        printf("Case #%lld: %lld\n",Case,maxx);
    }
    return 0;
}

題目連結：L題

題目大意：有四個隊伍進行足球比賽，贏方得3分，敗方不得分，平局雙方各得一分，給出最後得分，問是否合法，或者是否有多種情況

題目思路：因為比賽就6場，直接dfs一下就好了

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

using namespace std;
typedef long long ll;

struct node{
    int a,b,c,d;
}res[1005];

int cnt = 0;

void dfs(int now,int a,int b,int c,int d){
    if(now == 1){
        dfs(now+1,a+3,b,c,d);
        dfs(now+1,a,b+3,c,d);
        dfs(now+1,a+1,b+1,c,d);
    }
    else if(now == 2){
        dfs(now+1,a+3,b,c,d);
        dfs(now+1,a,b,c+3,d);
        dfs(now+1,a+1,b,c+1,d);
    }
    else if(now == 3){
        dfs(now+1,a+3,b,c,d);
        dfs(now+1,a,b,c,d+3);
        dfs(now+1,a+1,b,c,d+1);
    }
    else if(now == 4){
        dfs(now+1,a,b+3,c,d);
        dfs(now+1,a,b,c+3,d);
        dfs(now+1,a,b+1,c+1,d);
    }
    else if(now == 5){
        dfs(now+1,a,b,c,d+3);
        dfs(now+1,a,b+3,c,d);
        dfs(now+1,a,b+1,c,d+1);
    }
    else if(now == 6){
        dfs(now+1,a,b,c,d+3);
        dfs(now+1,a,b,c+3,d);
        dfs(now+1,a,b,c+1,d+1);
    }
    else if(now == 7){
        res[cnt].a = a;
        res[cnt].b = b;
        res[cnt].c = c;
        res[cnt].d = d;
        cnt++;
        return ;
    }
}


int main(){
    int T,a,b,c,d;
    scanf("%d",&T);
    dfs(1,0,0,0,0);
    for(int Case = 1;Case <= T;Case++){
        scanf("%d%d%d%d",&a,&b,&c,&d);
        int cot = 0;
        for(int i = 0;i < cnt;i++){
            if(res[i].a == a&&res[i].b == b&&res[i].c == c&&res[i].d == d) cot++;
        }
        if(cot == 0) printf("Case #%d: Wrong Scoreboard\n",Case);
        else if(cot == 1) printf("Case #%d: Yes\n",Case);
        else printf("Case #%d: No\n",Case);
    }
    return 0;
}