Educational Codeforces Round 50 (ABCD)

A. Function Height

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a set of $\mathrm{2n+12n+1 integer\; points\; on\; a\; Cartesian\; plane.\; Points\; are\; numbered\; from 00 to 2n2n inclusive.\; Let PiPi be\; the ii-th\; point.\; The xx-coordinate\; of\; the\; point PiPi equals ii.\; The yy-coordinate\; of\; the\; point PiPi equals\; zero\; (initially).\; Thus,\; initially Pi=(i,0)Pi=(i,0).}$

The given points are vertices of a plot of a piecewise function. The $\mathrm{jj-th\; piece\; of\; the\; function\; is\; the\; segment PjPj+1PjPj+1.}$

In one move you can increase the $\mathrm{yy-coordinate\; of\; any\; point\; with\; odd xx-coordinate\; (i.e.\; such\; points\; are P1,P3,\dots ,P2n-1P1,P3,\dots ,P2n-1)\; by 11.\; Note\; that\; the\; corresponding\; segments\; also\; change.}$

For example, the following plot shows a function for $\mathrm{n=3n=3 (i.e.\; number\; of\; points\; is 2\cdot 3+1=72\cdot 3+1=7)\; in\; which\; we\; increased\; the yy-coordinate\; of\; the\; point P1P1 three\; times\; and yy-coordinate\; of\; the\; point P5P5 one\; time:}$

Let the area of the plot be the area below this plot and above the coordinate axis OX. For example, the area of the plot on the picture above is 4 (the light blue area on the picture above is the area of the plot drawn on it).

Let the height of the plot be the maximum $\mathrm{yy-coordinate\; among\; all\; initial\; points\; in\; the\; plot\; (i.e.\; points P0,P1,\dots ,P2nP0,P1,\dots ,P2n).\; The\; height\; of\; the\; plot\; on\; the\; picture\; above\; is\; 3.}$

Your problem is to say which minimum possible height can have the plot consisting of $\mathrm{2n+12n+1 vertices\; and\; having\; an\; area\; equal\; to kk.\; Note\; that\; it\; is\; unnecessary\; to\; minimize\; the\; number\; of\; moves.}$

It is easy to see that any answer which can be obtained by performing moves described above always exists and is an integer number not exceeding ${10181018.}^{}$

Input

The first line of the input contains two integers $\mathrm{nn and kk (1\le n,k\le 10181\le n,k\le 1018)\; —\; the\; number\; of\; vertices\; in\; a\; plot\; of\; a\; piecewise\; function\; and\; the\; area\; we\; need\; to\; obtain.}$

Output

Print one integer — the minimum possible height of a plot consisting of $\mathrm{2n+12n+1 vertices\; and\; with\; an\; area\; equals kk.\; It\; is\; easy\; to\; see\; that\; any\; answer\; which\; can\; be\; obtained\; by\; performing\; moves\; described\; above\; always\; exists\; and\; is\; an\; integer\; number\; not\; exceeding 10181018.}$

Examples

input

Copy

4 3

output

Copy

1

input

Copy

4 12

output

Copy

3

input

Copy

999999999999999999 999999999999999986

output

Copy

1

Note

One of the possible answers to the first example:

The area of this plot is 3, the height of this plot is 1.

There is only one possible answer to the second example:

The area of this plot is 12, the height of this plot is 3.

#include <bits/stdc++.h>
#define ll long long int
using namespace std;

ll n,m;
int main(){
    cin>>n>>m;
    ll ans = m/n;
    ans += m%n==0?0:1;
    cout<<ans<<endl;
    return 0;
}

B. Diagonal Walking v.2

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Mikhail walks on a Cartesian plane. He starts at the point $(0,0)(0,0),\; and\; in\; one\; move\; he\; can\; go\; to\; any\; of\; eight\; adjacent\; points.\; For\; example,\; if\; Mikhail\; is\; currently\; at\; the\; point (0,0)(0,0),\; he\; can\; go\; to\; any\; of\; the\; following\; points\; in\; one\; move:$

• $(1,0)(1,0);$
• $(1,1)(1,1);$
• $(0,1)(0,1);$
• $(-1,1)(-1,1);$
• $(-1,0)(-1,0);$
• $(-1,-1)(-1,-1);$
• $(0,-1)(0,-1);$
• $(1,-1)(1,-1).$

If Mikhail goes from the point $(x1,y1)(x1,y1) to\; the\; point (x2,y2)(x2,y2) in\; one\; move,\; and x1\ne x2x1\ne x2 and y1\ne y2y1\ne y2,\; then\; such\; a\; move\; is\; called\; a diagonal\; move.$

Mikhail has $\mathrm{qq queries.\; For\; the ii-th\; query\; Mikhail\text{'}s\; target\; is\; to\; go\; to\; the\; point (ni,mi)(ni,mi) from\; the\; point (0,0)(0,0) in exactly kiki moves.\; Among\; all\; possible\; movements\; he\; want\; to\; choose\; one\; with\; the\; maximum\; number\; of diagonal\; moves.\; Your\; task\; is\; to\; find\; the\; maximum\; number\; of diagonal\; moves or\; find\; that\; it\; is\; impossible\; to\; go\; from\; the\; point (0,0)(0,0) to\; the\; point (ni,mi)(ni,mi) in kiki moves.}$

Note that Mikhail can visit any point any number of times (even the destination point!).

Input

The first line of the input contains one integer $\mathrm{qq (1\le q\le 1041\le q\le 104)\; —\; the\; number\; of\; queries.}$

Then $\mathrm{qq lines\; follow.\; The ii-th\; of\; these qq lines\; contains\; three\; integers nini, mimi and kiki (1\le ni,mi,ki\le 10181\le ni,mi,ki\le 1018)\; — xx-coordinate\; of\; the\; destination\; point\; of\; the\; query, yy-coordinate\; of\; the\; destination\; point\; of\; the\; query\; and\; the\; number\; of\; moves\; in\; the\; query,\; correspondingly.}$

Output

Print $\mathrm{qq integers.\; The ii-th\; integer\; should\; be\; equal\; to -1 if\; Mikhail\; cannot\; go\; from\; the\; point (0,0)(0,0) to\; the\; point (ni,mi)(ni,mi) in exactly kiki moves\; described\; above.\; Otherwise\; the ii-th\; integer\; should\; be\; equal\; to\; the\; the\; maximum\; number\; of diagonal\; moves among\; all\; possible\; movements.}$

Example

input

Copy

3
2 2 3
4 3 7
10 1 9

output

Copy

1
6
-1

Note

One of the possible answers to the first test case: $(0,0)\to (1,0)\to (1,1)\to (2,2)(0,0)\to (1,0)\to (1,1)\to (2,2).$

One of the possible answers to the second test case: $(0,0)\to (0,1)\to (1,2)\to (0,3)\to (1,4)\to (2,3)\to (3,2)\to (4,3)(0,0)\to (0,1)\to (1,2)\to (0,3)\to (1,4)\to (2,3)\to (3,2)\to (4,3).$

In the third test case Mikhail cannot reach the point $(10,1)(10,1) in\; 9\; moves.$

仔细分析就出来了。

#include <bits/stdc++.h>
#define ll long long int
using namespace std;

int main(){
    ll t,x,y,z;
    cin>>t;
    while(t--){
        cin>>x>>y>>z;
        ll sum = min(x,y);
        ll ans = abs(x-y);
        if(z<max(x,y)){
            cout<<"-1"<<endl;
        }else{
            if(ans%2){
                cout<<z-1<<endl;
            }else{
                if(z==sum+ans){
                    cout<<z<<endl;
                    continue;
                }
                ll cnt = z-sum-ans;
                if(cnt%2==0){
                    cout<<z<<endl;
                }else{
                    cout<<z-2<<endl;
                }
            }
        }
    }
    return 0;
}

C. Classy Numbers

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Let's call some positive integer classy if its decimal representation contains no more than $\mathrm{33 non-zero\; digits.\; For\; example,\; numbers 44, 200000200000, 1020310203 are classy and\; numbers 42314231, 102306102306, 72774200007277420000 are\; not.}$

You are given a segment $[L;R][L;R].\; Count\; the\; number\; of classy integers xx such\; that L\le x\le RL\le x\le R.$

Each testcase contains several segments, for each of them you are required to solve the problem separately.

Input

The first line contains a single integer $\mathrm{TT (1\le T\le 1041\le T\le 104)\; —\; the\; number\; of\; segments\; in\; a\; testcase.}$

Each of the next $\mathrm{TT lines\; contains\; two\; integers LiLi and RiRi (1\le Li\le Ri\le 10181\le Li\le Ri\le 1018).}$

Output

Print $\mathrm{TT lines\; —\; the ii-th\; line\; should\; contain\; the\; number\; of classy integers\; on\; a\; segment [Li;Ri][Li;Ri].}$

Example

input

Copy

4
1 1000
1024 1024
65536 65536
999999 1000001

output

Copy

1000
1
0
2

数位dp （看了好久才学会）
算是模板题吧

#include <bits/stdc++.h>
#define ll long long int
using namespace std;
ll dp[30][4],digth[30];


ll dfs(int len, int num,  bool lim){
    if(len == 0)
        return 1;
    if(!lim && dp[len][num] != -1)
        return dp[len][num];
    ll ans = 0;
    int up = lim ? digth[len] : 9;

    for(int i = 0; i <= up; ++i){
        if(num < 3){
            if( !i )
                ans += dfs(len - 1, num, lim && i == digth[len] );
            else
                ans += dfs(len - 1, num + 1, lim && i == digth[len] );
        }else if(num == 3){
            if(!i)
                ans += dfs(len -1, num, lim && i == digth[len] );
        }
    }
    if(!lim )
        dp[len][num] = ans;
    return ans;
}

ll solve(ll x){
    int k = 0;
    while(x){
        digth[ ++k ] = x%10;
        x /= 10;
    }
    return dfs(k,0,true);
}

int t;
ll n,m;
int main(){
    scanf("%d", &t);
    memset( dp , -1 , sizeof(dp) ) ;
    while(t--){
        scanf("%lld%lld",&n,&m);
        // printf("%lld %lld
",solve(m),solve(n-1));
        printf("%lld
", solve(m)-solve(n-1));        
    }
    return 0;
}

D. Vasya and Arrays

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya has two arrays $\mathrm{AA and BB of\; lengths nn and mm,\; respectively.}$

He can perform the following operation arbitrary number of times (possibly zero): he takes some consecutive subsegment of the array and replaces it with a single element, equal to the sum of all elements on this subsegment. For example, from the array $[1,10,100,1000,10000][1,10,100,1000,10000] Vasya\; can\; obtain\; array [1,1110,10000][1,1110,10000],\; and\; from\; array [1,2,3][1,2,3] Vasya\; can\; obtain\; array [6][6].$

Two arrays $\mathrm{AA and BB are\; considered\; equal\; if\; and\; only\; if\; they\; have\; the\; same\; length\; and\; for\; each\; valid ii Ai=BiAi=Bi.}$

Vasya wants to perform some of these operations on array $\mathrm{AA,\; some\; on\; array BB,\; in\; such\; a\; way\; that\; arrays AA and BB become\; equal.\; Moreover,\; the\; lengths\; of\; the\; resulting\; arrays\; should\; be\; maximal\; possible.}$

Help Vasya to determine the maximum length of the arrays that he can achieve or output that it is impossible to make arrays $\mathrm{AA and BBequal.}$

Input

The first line contains a single integer $\mathrm{n (1\le n\le 3\cdot 105)n (1\le n\le 3\cdot 105) —\; the\; length\; of\; the\; first\; array.}$

The second line contains $\mathrm{nn integers a1,a2,\cdots ,an (1\le ai\le 109)a1,a2,\cdots ,an (1\le ai\le 109) —\; elements\; of\; the\; array AA.}$

The third line contains a single integer $\mathrm{m (1\le m\le 3\cdot 105)m (1\le m\le 3\cdot 105) —\; the\; length\; of\; the\; second\; array.}$

The fourth line contains $\mathrm{mm integers b1,b2,\cdots ,bm (1\le bi\le 109)b1,b2,\cdots ,bm (1\le bi\le 109) -\; elements\; of\; the\; array BB.}$

Output

Print a single integer — the maximum length of the resulting arrays after some operations were performed on arrays $\mathrm{AA and BB in\; such\; a\; way\; that\; they\; became\; equal.}$

If there is no way to make array equal, print "-1".

Examples

input

Copy

5
11 2 3 5 7
4
11 7 3 7

output

Copy

3

input

Copy

2
1 2
1
100

output

Copy

-1

input

Copy

3
1 2 3
3
1 2 3

output

Copy

3

指针模拟

#include <bits/stdc++.h>
#define ll long long int
#define N 300005
using namespace std;
int n,m;
ll an[N],bn[N];
int main(){
    scanf("%d",&n);
    for(int i = 1; i <= n; ++i){
        scanf("%lld", &an[i]);
        an[i] += an[i-1];
    }
    scanf("%d", &m);
    for(int i = 1; i <= m; ++i){
        scanf("%lld", &bn[i]);
        bn[i] += bn[i-1];
    }
    if(an[n]!=bn[m]){
        printf("-1
");
        return 0;
    }
    int pox = 0;
    for(int i = 1,j = 1;i <= n&&j <= m;){
        if(an[i] == bn[j]){
            pox++;
            i++;j++;
        }else if(an[i] < bn[j]){
            i++;
        }else{
            j++;
        }
    }
    printf("%d
", pox);
    return 0;
}