codeforces - 932 (div2)

A.将给的字符串正反输出两边

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double eps = 1e-9;
const int maxn = 10010;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,M,K;
char str[maxn];
int main(){
    scanf("%s",str);
    int l = strlen(str);
    cout << str;
    for(int i = 0 ; i < l / 2; i ++) swap(str[i],str[l - i - 1]);
    cout << str << endl;
    return 0;
}
A

 

B.暴力递推出所有数字对应的最终值,求每个最终值得前缀和,查询的时候O(1)输出

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double eps = 1e-9;
const int maxn = 1e6 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,M,K;
int fa[maxn];
int pre[10][maxn];
int cul(int x){
    int ans = 1;
    while(x){
        if(x % 10) ans *= x % 10;
        x /= 10;
    }
    return ans;
}
void init(){
    for(int i = 1 ; i < 10; i ++){
        fa[i] = i;
    }
    for(int i = 10; i < maxn; i ++){
        fa[i] = fa[cul(i)];
    }
    for(int i = 1; i < maxn; i ++){
        for(int j = 1; j <= 9; j ++){
            pre[j][i] = pre[j][i - 1];
        }
        pre[fa[i]][i]++;
    }
}
int main(){
    init();
    Sca(N);
    while(N--){
        int l,r,k; Sca3(l,r,k);
        Pri(pre[k][r] - pre[k][l - 1]);
    }
    return 0;
}
B

 

C.主要在搞明白题意

先找出形如Ax + By = N,的正数x,y,N较小可以直接暴力

然后说明序列形成了x个长度A的环和y个长度B的环,构造输出就可以了

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double eps = 1e-9;
const int maxn = 1e6 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
LL N;
LL exgcd(LL a,LL b,LL &x,LL& y){
    if(!a && !b) return -1;
    if(!b){
        x = 1; y = 0;
        return a;
    }
    LL d = exgcd(b,a % b,y,x);
    y -= a / b * x;
    return d;
}
void solve(int &cnt,LL x,LL A){
    for(int i = 1; i <= x; i ++){
        int p = cnt;
        for(int j = 1; j < A; j ++){
            cnt++;
            printf("%d ",cnt + 1);
        }
        cnt++;
        printf("%d ",p + 1);
    }
}
LL A,B;
int main(){
    scanf("%lld%lld%lld",&N,&A,&B);
    LL x,y;
    bool flag = 0;
    for(int i = 0; i * A <= N ; i ++){
        if(!((N - i * A) % B)){
            x = i;
            y = (N - i * A) / B;
            flag = 1;
            break;
        }
    }
    if(!flag){puts("-1"); return 0;}
    int cnt = 0;
    solve(cnt,x,A);
    solve(cnt,y,B);
    return 0;
}
C

 

D.暴力肯定是不行的,极端数据是直接给一条从祖先到叶子权值递减的链,时间复杂度(2e5 * 2e5)

所以查询祖先要用树上倍增,时间复杂度nlogn

树上倍增维护每个节点的父亲值。

1.插入的时候如果当前节点的权值比给定的父亲节点权值小,就直接连上去

2.如果比给定的权值大,就倍增查找出他父亲的最小的比他权值大的祖先,将这个结点连到他祖先上去。

3.维护每个结点到根节点的一个前缀和

4.查询的之后同2,如果倍增查找到祖先前缀和-结点前缀和小于y的值,然后跳到y并计入两者深度差的贡献。

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double eps = 1e-9;
const int maxn = 4e5 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int SP = 20; 
int N,M,K;
int fa[maxn][SP],dep[maxn];
LL val[maxn],pre[maxn];
int main(){
    int Q; Sca(Q);
    LL la = 0;
    int cnt = 1;
    val[0] = 2e18; dep[1] = 1;
    while(Q--){
        LL op,x,y;
        scanf("%lld%lld%lld",&op,&x,&y);
        x ^= la; y ^= la;
    //    cout <<"bug" <<op << " " << x << " " << y << endl;
        if(op == 1){
            cnt++; val[cnt] = y;
            if(val[cnt] > val[x]){
                for(int i = SP - 1; i >= 0 ; i --){
                    int f = fa[x][i];
                    if(val[f] < val[cnt]) x = f; 
                }
                x = fa[x][0];
            }
            dep[cnt] = dep[x] + 1;
            fa[cnt][0] = x;
            pre[cnt] = pre[x] + val[cnt];
            for(int i = 1;i < SP; i ++) fa[cnt][i] = fa[fa[cnt][i - 1]][i - 1];
         }else{
             if(y < val[x]){
                 la = 0; puts("0");
                 continue;
            }
            y -= val[x];
            la = 1;
            x = fa[x][0];
             for(int i = SP - 1; i >= 0 && x != 0; i --){
                 int f = fa[x][i];
                 LL p = pre[x] - pre[f];
                 if(y >= p){
                     y -= p;
                     la += (dep[x] - dep[f]);
                     x = f;
                 }
             }
             Prl(la); //la
         }
    }
    return 0;
}
D

 

posted @ 2019-07-17 17:02  Hugh_Locke  阅读(269)  评论(0编辑  收藏  举报