AIM Tech Round 3 (Div. 2) 题解

总结：做了3道，补题1道

A. Juicer

题意：水题

#include <iostream>
#include <cstdio>
#include <algorithm>
#define maxn 100005
using namespace std;
int a[maxn];
int main()
{
    int n,b,d;

    cin >> n >> b >> d;

    for(int i = 0;i < n;i ++) scanf("%d",&a[i]);
    int sum = 0;
    int ans = 0;
    for(int i = 0;i < n;i ++){
        if(a[i] > b) continue;
        else{
            sum += a[i];
            if(sum > d){
                ans ++;
                sum = 0;
            }
        }
    }
    cout << ans << endl;
    return 0;
}

B. Checkpoints

题意：水题

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <map>
#include <cstring>
#define maxn 100005
using namespace std;
typedef long long ll;
int main()
{int left[maxn];
int right[maxn];
int lcnt,rcnt;
    map<int,int> mmap;
    int n,a;
    cin >> n >> a;
    ll ans;
    lcnt = rcnt = 0;
    for(int i = 0;i < n;i ++){
        int d;
        scanf("%d",&d);
        mmap[d] ++;
        if(d < a){
            left[lcnt++] = d;
        }
        else if(d > a){
            right[rcnt++] = d;
        }
    }
    sort(left,left+lcnt);
    sort(right,right+rcnt);
    if(lcnt + rcnt <= 1) cout << "0" << endl;
    else if(lcnt == 0 || rcnt == 0){
        if(lcnt == 0){
            if(mmap[right[rcnt-1]] > 1) ans = right[rcnt-1] - a;
            else ans = right[rcnt-2] - a;
        }
        else{
            if(mmap[left[0]] > 1) ans = (a-left[0]);
            else ans = (a-left[1]);
        }
        cout << ans << endl;
    }
    else{
        if(mmap[left[0]] > 1 && mmap[right[rcnt-1]] > 1) {
            ans = max(a-left[0],right[rcnt-1]-a) + 2*min(a-left[0],right[rcnt-1]-a);
        }
        else if(mmap[left[0]] > 1 && mmap[right[rcnt-1]] == 1){
            ll t1,t2;
            t1 = (a-left[0]) + 2*(right[rcnt-2] - a);
            t2 = 2*(a-left[0]) + (right[rcnt-2] - a);
            ans = min(t1,t2);
        }
        else if(mmap[left[0]] == 1 && mmap[right[rcnt-1]] > 1){
            ll t1,t2;
            t1 = 2*(a-left[1]) + (right[rcnt-1] - a);
            t2 = (a-left[1]) + 2*(right[rcnt-1] - a);
            ans = min(t1,t2);
        }
        else{
            ll t1,t2,t3,t4;
            if(rcnt > 1){
                t1 = (a-left[0]) + 2*(right[rcnt-2] - a);
                t3 = 2*(a-left[0]) + (right[rcnt-2] - a);
            }
            else{
                t1 = (a-left[0]) + 0;
                t3 = 2*(a-left[0]) + 0;
            }
            if(lcnt > 1){
                t2 = 2*(a-left[1]) + (right[rcnt-1] - a);
                t4 = (a - left[1]) + 2*(right[rcnt-1] - a);
            }
            else{
                t2 = 0 + (right[rcnt-1] - a);
                t4 = 0 + 2*(right[rcnt-1] - a);
            }
            t1 = min(t1,t2);
            t3 = min(t3,t4);
            ans = min(t1,t3);
        }
        cout << ans << endl;
    }
    return 0;
}

C. Letters Cyclic Shift

水题

#include <iostream>
#include <string>
using namespace std;

int main()
{
    string s;
    int i,k,t;
    cin >> s;
    for(k = 0;k < s.length();k ++){
        if(s[k] != 'a') break;
    }
    for(t = k;t < s.length();t ++){
        if(s[t] == 'a') break;
    }
    for(i = k;i < t;i ++){
        if(s[i] == 'a') s[i] = 'z';
        else s[i] --;
    }
    if(k == s.length()) s[k-1] = 'z';
    cout << s << endl;
    return 0;
}

D. Recover the String

题意：给出四个数a0，a1，a2，a3，求一个只有0、1的数字串，其子串（不一定连续）“00”的个数是a0，“01”的个数是a1，以此类推。

思路:首先由a0,a3可以推出数字串中0,1的个数分别为s0，s1，则a0+a1+a2+a3=c(s0+s1,2)。如果初始为所有0在所有1前面，将一个1左移到一个0前面，则10增加1,01减少1。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <map>
#include <cstring>
using namespace std;
typedef long long ll;
#define maxn 1000005
ll c[maxn];
void init(){
    c[0] = c[1] = 0;
    for(ll i = 2;i < maxn;i ++){
        c[i] = (i * (i - 1)) / 2;
    }
}
int main()
{
    ll a0,a1,a2,a3;

    cin >> a0 >> a1 >> a2 >> a3;
    if(a0 + a1 + a2 + a3 == 0) {
        cout << "0" << endl;
        return 0;
    }
    init();
    int s0,s1;         //代表0,1的个数
    bool flag1,flag2;
    flag1 = flag2 = false;
    for(int i = 0;i < maxn;i ++){
        if(c[i] == a0){
            s0 = i;
            flag1 = true;
            break;
        }
    }
    for(int i = 0;i < maxn;i ++){
        if(c[i] == a3){
            s1 = i;
            flag2 = true;
            break;
        }
    }
    if(!flag1 || !flag2){
        cout << "Impossible" << endl;
    }
    else{
        if(s1 == 0){
            if(a1 || a2) s1 = 1;
        }
        if(s0 == 0){
            if(a1 || a2) s0 = 1;
        }
        if((a1+a2+a3+a0) != c[s1+s0]) cout << "Impossible" << endl;
        else if(!s0 || !s1){
            if(s0) for(int i = 0;i < s0;i ++) cout << "0";
            if(s1) for(int i = 0;i < s1;i ++) cout << "1";
            cout << endl;
        }
        else{
            int s = a2;
            int cc = s/s0;
            int dd = s0 - s % s0;            //s0可能是0，所以前面分类讨论要把s0==0 || s1 == 0的case单独讨论
            for(int i = 0;i < cc;i ++) cout << "1";
            for(int i = 0;i < dd;i ++) cout << "0";
            if(s%s0)
                cout << "1";
            for(int i = 0;i < s % s0;i ++) cout << "0";
            int ee = s1-s/s0;
            ee = s%s0 == 0? ee : ee - 1;
            for(int i = 0;i < ee;i ++) cout << "1";
            cout << endl;
        }
    }
    return 0;
}