2016-2017 ACM-ICPC CHINA-Final

题目链接：A题

题目大意：问有多少个小于2n的形式是2k−1的能整除7的数字的个数

题目思路：7的二进制是111，而2k−1的二进制是k个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;}

题目链接：D题

题目大意：有n个蛋糕，每个蛋糕都有尺寸，现在问你能形成多少个蛋糕塔，要求：蛋糕塔的层数一定得是k，每一层只有能一个蛋糕，下一层的蛋糕得尺寸至少得是上一层得两倍

题目思路：二分蛋糕塔的个数，然后我们优先取最小的mid个，然后线性扫看能合成多少层，然后层数和k比较得二分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;}

题目链接：L题

题目大意：有四个队伍进行足球比赛，赢方得3分，败方不得分，平局双方各得一分，给出最后得分，问是否合法，或者是否有多种情况

题目思路：因为比赛就6场，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;}