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  • CodeForces Round#567 Div.2

    A. Chunga-Changa

    #include <bits/stdc++.h>
    using namespace std;
     
    long long x, y, z;
     
    int main() {
        long long ans = 0, cnt = 0;
        scanf("%lld%lld%lld", &x, &y, &z);
        ans = (x + y) / z;
        long long p1 = x % z, p2 = y % z;
        if(z - p1 < z - p2) {
            if(p2 >= z - p1) cnt = z - p1;
        } else {
            if(p1 >= z - p2) cnt = z - p2;
        }
        printf("%lld %lld
    ", ans, cnt);
        return 0;
    }
    View Code

    B. Split a Number

    By Golden_miner, contest: Codeforces Round #567 (Div. 2), problem: (B) Split a Number, Accepted, #
    #include <bits/stdc++.h>
    using namespace std;
     
    int N;
    string s, ans, temp;
     
    string add(string a, string b) {
        string c;
        int len1 = a.length();
        int len2 = b.length();
        int len = max(len1, len2);
        for(int i = len1; i < len; i ++)
            a = "0" + a;
        for(int i = len2; i < len; i ++)
            b= "0" + b;
        int ok = 0;
        for(int i = len - 1; i >= 0; i --) {
            char temp = a[i] + b[i] - '0' + ok;
            if(temp > '9') {
                ok = 1;
                temp -= 10;
            }
            else ok = 0;
            c = temp + c;
        }
        if(ok) c = "1" + c;
        return c;
    }
     
    string DelZero(string t) {
        string ans = "";
        int temp = 0;
        for(int i = 0; t[i]; i ++) {
            if(t[i] != '0') {
                temp = i;
                break;
            }
        }
        ans = t.substr(temp, t.length() - temp);
        return ans;
    }
     
    string Sum(int x) {
        string num1 = s.substr(0, x);
        string num2 = s.substr(x, N - x);
        string ans = add(num1, num2);
        return ans;
    }
     
    bool Bigger(string &a, string &b) {
        if(a.length() > b.length()) return true;
        else if(a.length() < b.length()) return false;
        else {
            int len = a.length();
            for(int i = 0; i < len; i ++) {
                if(a[i] == b[i]) continue;
                if(a[i] > b[i]) return true;
                return false;
            }
        }
    }
     
    int main() {
        scanf("%d", &N);
        cin >> s;
     
        ans = s;
     
        int pos = (N - 1) / 2, p = pos;
        int cnt = 0;
        while(p >= 0 && cnt <= 1) {
            if(s[p + 1] != '0') {
                temp = Sum(p + 1);
                if(Bigger(ans, temp)) ans = temp;
                cnt ++;
            }
            p --;
        }
        p = pos;
     
        cnt = 0;
        while(p < N - 1 && cnt <= 0) {
            if(s[p + 1] != '0') {
                temp = Sum(p + 1);
                if(Bigger(ans, temp)) ans = temp;
                cnt ++;
            }
            p ++;
        }
     
        for(int i = 0; i < ans.length();i ++) {
            printf("%c", ans[i]);
        }
     
     
        return 0;
    }
    View Code

    C. Flag

    #include <bits/stdc++.h>
    using namespace std;
     
    char mp[1010][1010];
    int dp[1010][1010];
    int N, M;
     
    int main() {
        scanf("%d%d", &N, &M);
        for(int i = 1; i <= N; i ++) 
            scanf("%s", mp[i] + 1);
        
        memset(dp, 0, sizeof(dp));
        for(int i = N; i >= 1; i --) {
            for(int j = 1; j <= M; j ++) 
                if(mp[i][j] == mp[i + 1][j])
                    dp[i][j] = dp[i + 1][j] + 1;
                else dp[i][j] = 1;
        }
        
        long long ans = 0;
        for(int i = 1; i <= N; i ++) {
            for(int j = 1, k = 0; j <= M; j ++) {
                int h = dp[i][j];
                if(i + 3 * h - 1 <= N && mp[i][j] != mp[i + h][j] && dp[i + h][j] == h
                  && mp[i + h][j] != mp[i + 2 * h][j] && dp[i + 2 * h][j] >= h) {
                    if(k && mp[i][j] == mp[i][j - 1] && dp[i][j - 1] == h
                      && mp[i + h][j] == mp[i + h][j - 1] && dp[i + h][j - 1] == h
                      && mp[i + 2 * h][j] == mp[i + 2 * h][j - 1] && dp[i + 2 * h][j - 1] >= h)
                        ++ k;
                    else k = 1;
                } else k = 0;
                
                ans += k;
                
            }
        }
        
        printf("%lld
    ", ans);
        
        return 0;
    }
    View Code

    D. Irrigation

    #include <bits/stdc++.h>
    using namespace std;
    
    #define MAX 500010
    #define LL long long
    #define MOD 1000000007
    
    struct node{
        int id;
        LL q;
    };
    
    bool cmp(node a,node b){
        return a.q<b.q;
    }
    
    node rr[MAX];
    int ans[MAX];
    int s[MAX*5];
    int a[MAX];
    vector<int> e[MAX];
    
    void update(int l,int r,int tt,int k){
        if(l==r){
            s[tt]++;
            return ;
        }
        int mid=(l+r)/2;
        if(mid<k)
            update(mid+1,r,tt*2+1,k);
        else
            update(l,mid,tt*2,k);
    
        s[tt]=s[tt*2]+s[tt*2+1];
    }
    
    int query(int l,int r,int tt,int k){
        if(l==r)
            return l;
        int mid=(l+r)/2;
        if(s[tt*2]<k)
            return query(mid+1,r,tt*2+1,k-s[tt*2]);
        else
            return query(l,mid,tt*2,k);
    }
    
    int main(){
    
        int n,m,q,x;
        scanf("%d%d%d",&n,&m,&q);
        for(int i=1;i<=n;i++){
            scanf("%d",&x);
            a[x]++;
        }
    
        LL st=n+1;
        int sz=0;
    
        for(int i=1;i<=m;i++){
            e[a[i]].push_back(i);
            st=min((LL)a[i],st);
        }
        sz=e[st].size();
        for(int i=0;i<e[st].size();i++){
            update(1,m,1,e[st][i]);
        }
    
        LL y,pre=n;
        for(int i=1;i<=q;i++){
            scanf("%lld",&y);
            rr[i].id=i;
            rr[i].q=y;
        }
        sort(rr+1,rr+1+q,cmp);
    
        for(int i=1;i<=q;i++){
            LL y=rr[i].q-pre;
    
            if(st>n){
                ans[rr[i].id]=(y-1)%m+1;
            }
            else{
                while(y>sz&&st<=n){
    
                    y-=sz;
                    pre+=sz;
                    st++;
                    sz+=e[st].size();
    
                    for(int j=0;j<e[st].size();j++){
                        update(1,m,1,e[st][j]);
                    }
                }
    
                if(st>n){
                    ans[rr[i].id]=(y-1)%m+1;
                }
                else
                    ans[rr[i].id]=query(1,m,1,y);
            }
        }
        printf("
    ");
        for(int i=1;i<=q;i++)
            printf("%d
    ",ans[i]);
        return 0;
    }
    View Code

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  • 原文地址:https://www.cnblogs.com/zlrrrr/p/11593285.html
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