2017 Multi-University Training Contest

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rank:77/758

1001 Rikka with Candies

因为没有手写bitset被卡了一下，最后过了。

#include <bits/stdc++.h>
const long long mod = 1e9+7;
const double ex = 1e-10;
#define inf 0x3f3f3f3f
using namespace std;
typedef pair<int,int>  pii;
bitset<50005> A;
bitset<50005> B;
bitset<50005> an;
vector <pii> E;
int ans[50005];
int ansid[50005];
int main()
{
    int t;
    scanf("%d",&t);
    int c;
    int maxa;
    while (t--){
        int N,M,Q;
        scanf("%d%d%d",&N,&M,&Q);
        A.reset();
        B.reset();
        maxa = 0;
        for (int i = 1; i<=N; i++){
            scanf("%d",&c);
            A[c] = 1;
            maxa = max(maxa,c);
        }
        E.clear();
        for (int i = 0; i<=maxa ;i++) ans[i] = -1;

        for (int i = 1;i<=M; i++){
            scanf("%d",&c);
            E.push_back(make_pair(c,-1));
        }


        for (int i = 1;i<=Q;i++){
            scanf("%d",&c);
            E.push_back(pii(c,i));
        }
        sort(E.begin(),E.end());
        for (int i = E.size()-1 ; i>=0; i--)
        {
            c = E[i].first;
            if (E[i].second == -1){
            for (int j = 0;j<=maxa;j+=c)
                B[j] = !B[j];
            }
            else{
                int id = E[i].second;
                if (c>maxa) ansid[id] = 0;
                else{
                    if (ans[c]!=-1) ansid[id] = ans[c];
                    else
                    {
                      an = (A>>c)&(B);
                      ans[c] = (an.count()) % 2;
                      ansid[id] = ans[c];
                    }
                }
            }
        }
        for (int i = 1;i<=Q;i++)
            printf("%d
",ansid[i]);
    }
    return 0;
}

1006

贪心构造

#include <bits/stdc++.h>
using namespace std;
int T;
long long n,m;
int main()
{
    scanf("%d",&T);
    while (T--) {
        scanf("%I64d%I64d",&n,&m);
        if (n==1) {
            puts("0");
            continue;
        }
        if (n==2) {
            puts("2");
            continue;
        }
        if (m<n-1) {
            printf("%I64d
",2LL*m+2LL*m*(m-1)+(n*(n-1)-2LL*m-m*(m-1))*n);
        } else if (m<n*(n-1)/2) {
            printf("%I64d
",2LL*m+(n*(n-1)-2*m)*2LL);
        } else {
            printf("%I64d
",n*(n-1));
        }
    }
    return 0;
}

1008

枚举，反向考虑每个元素对之后的贡献。

#include <bits/stdc++.h>
const long long mod = 1e9+7;
const double ex = 1e-10;
#define inf 0x3f3f3f3f
using namespace std;
long long M[10005];
long long TM[10005];
int main()
{
    int T;
    scanf("%d",&T);
    while (T--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        for (int i = 0; i<=m; i++)
             TM[i] = 0;
        for (int i= 0; i<=m;i++)
        {
            long long z;
            scanf("%I64d",&z);
            M[i] = z;
        }
        int tmp = 0;
        TM[0] = 1;
        for (int i = 1; i<=n ;){
            if (M[tmp]>TM[tmp])
            {
                printf("%d%c",tmp,i==n?'
':' ');
                for (int j = m; j>=0;j--){
                    TM[j+tmp] = TM[j+tmp] + TM[j];
                }
                i++;
            }
            else
            {
                tmp++;
            }
        }
    }
    return 0;
}

1011

倒着递推一波即可。

#include <bits/stdc++.h>
#define maxn 100010
#define inf 0x3f3f3f3f
#define REP(i,x,y) for(int i=x;i<(y);i++)
#define RREP(i,x,y) for(int i=x;i>(y);i--)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
int n,k,a[maxn],win[maxn];
set<int>S;
int main()
{
    int T;scanf("%d",&T);
    while(T--) {
        S.clear();
        memset(win,0,sizeof(win));
        scanf("%d %d",&n,&k);
        REP(i,1,n+1) {
            scanf("%d",&a[i]);
        }
        sort(a+1,a+1+n);
        int ans=1;
        win[n]=1;
        RREP(i,n-1,0) {
            if(win[i+1]==1&&a[i+1]-a[i]<=k) {
                win[i]=1;
                ans++;
            }
        }
        cout<<ans<<endl;
    }
}