Codeforces Round #501 (Div. 3) ABDE1E2

A. Points in Segments

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a set of $\mathrm{nn segments\; on\; the\; axis OxOx,\; each\; segment\; has\; integer\; endpoints\; between 11 and mm inclusive.\; Segments\; may\; intersect,\; overlap\; or\; even\; coincide\; with\; each\; other.\; Each\; segment\; is\; characterized\; by\; two\; integers lili and riri (1\le li\le ri\le m1\le li\le ri\le m)\; —\; coordinates\; of\; the\; left\; and\; of\; the\; right\; endpoints.}$

Consider all integer points between $\mathrm{11 and mm inclusive.\; Your\; task\; is\; to\; print\; all\; such\; points\; that\; don\text{'}t\; belong\; to\; any\; segment.\; The\; point xxbelongs\; to\; the\; segment [l;r][l;r] if\; and\; only\; if l\le x\le rl\le x\le r.}$

Input

The first line of the input contains two integers $\mathrm{nn and mm (1\le n,m\le 1001\le n,m\le 100)\; —\; the\; number\; of\; segments\; and\; the\; upper\; bound\; for\; coordinates.}$

The next $\mathrm{nn lines\; contain\; two\; integers\; each lili and riri (1\le li\le ri\le m1\le li\le ri\le m)\; —\; the\; endpoints\; of\; the ii-th\; segment.\; Segments\; may\; intersect,\; overlap\; or\; even\; coincide\; with\; each\; other.\; Note,\; it\; is\; possible\; that li=rili=ri,\; i.e.\; a\; segment\; can\; degenerate\; to\; a\; point.}$

Output

In the first line print one integer $\mathrm{kk —\; the\; number\; of\; points\; that\; don\text{'}t\; belong\; to\; any\; segment.}$

In the second line print exactly $\mathrm{kk integers\; in any order\; —\; the\; points\; that\; don\text{'}t\; belong\; to\; any\; segment.\; All\; points\; you\; print\; should\; be\; distinct.}$

If there are no such points at all, print a single integer $\mathrm{00 in\; the\; first\; line\; and\; either\; leave\; the\; second\; line\; empty\; or\; do\; not\; print\; it\; at\; all.}$

Examples

input

Copy

3 5
2 2
1 2
5 5

output

Copy

2
3 4 

input

Copy

1 7
1 7

output

Copy

0

Note

In the first example the point $\mathrm{11 belongs\; to\; the\; second\; segment,\; the\; point 22 belongs\; to\; the\; first\; and\; the\; second\; segments\; and\; the\; point 55belongs\; to\; the\; third\; segment.\; The\; points 33 and 44 do\; not\; belong\; to\; any\; segment.}$

In the second example all the points from $\mathrm{11 to 77 belong\; to\; the\; first\; segment.}$

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e5+10;
int a[N];
int main() {
    int n, m, l, r;
    cin >> n >> m;
    for(int i = 1; i <= n; i ++) {
        cin >> l >> r;
        for(int j = l; j <= r; j ++) a[j] = 1;
    }
    int ans = 0;
    for(int i = 1; i <= m; i ++) if(!a[i]) ans++;
    printf("%d
",ans);
    for(int i = 1; i <= m; i ++) if(!a[i]) printf("%d ",i);
    return 0;
}

B. Obtaining the String

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two strings $\mathrm{ss and tt.\; Both\; strings\; have\; length nn and\; consist\; of\; lowercase\; Latin\; letters.\; The\; characters\; in\; the\; strings\; are\; numbered\; from 11 to nn.}$

You can successively perform the following move any number of times (possibly, zero):

• swap any two adjacent (neighboring) characters of $\mathrm{ss (i.e.\; for\; any i=\{1,2,\dots ,n-1\}i=\{1,2,\dots ,n-1\} you\; can\; swap sisi and si+1)si+1).}$

You can't apply a move to the string $\mathrm{tt.\; The\; moves\; are\; applied\; to\; the\; string ss one\; after\; another.}$

Your task is to obtain the string $\mathrm{tt from\; the\; string ss.\; Find\; any\; way\; to\; do\; it\; with\; at\; most 104104 such\; moves.}$

You do not have to minimize the number of moves, just find any sequence of moves of length ${\mathrm{104104 or\; less\; to\; transform ss into tt.}}^{}$

Input

The first line of the input contains one integer $\mathrm{nn (1\le n\le 501\le n\le 50)\; —\; the\; length\; of\; strings ss and tt.}$

The second line of the input contains the string $\mathrm{ss consisting\; of nn lowercase\; Latin\; letters.}$

The third line of the input contains the string $\mathrm{tt consisting\; of nn lowercase\; Latin\; letters.}$

Output

If it is impossible to obtain the string $\mathrm{tt using\; moves,\; print\; "-1".}$

Otherwise in the first line print one integer $\mathrm{kk —\; the\; number\; of\; moves\; to\; transform ss to tt.\; Note\; that kk must\; be\; an\; integer\; number\; between 00and 104104 inclusive.}$

In the second line print $\mathrm{kk integers cjcj (1\le cj<n1\le cj<n),\; where cjcj means\; that\; on\; the jj-th\; move\; you\; swap\; characters scjscj and scj+1scj+1.}$

If you do not need to apply any moves, print a single integer $\mathrm{00 in\; the\; first\; line\; and\; either\; leave\; the\; second\; line\; empty\; or\; do\; not\; print\; it\; at\; all.}$

Examples

input

Copy

6
abcdef
abdfec

output

Copy

4
3 5 4 5 

input

Copy

4
abcd
accd

output

Copy

-1

Note

In the first example the string $\mathrm{ss changes\; as\; follows:\; "abcdef" \to \to "abdcef" \to \to "abdcfe" \to \to "abdfce" \to \to "abdfec".}$

In the second example there is no way to transform the string $\mathrm{ss into\; the\; string tt through\; any\; allowed\; moves.}$

$\mathrm{模拟题，将S变成t}$

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 110;
char s1[N], s2[N], s3[N];
int a[N], b[N];
int main() {
    int n, j;
    cin >> n >> s1 >> s2;
    for(int i = 0; i < n; i ++) a[s1[i]-'a']++, s3[i] = s1[i];
    for(int i = 0; i < n; i ++) b[s2[i]-'a']++;
    for(int i = 0; i < 26; i ++) {
        if(a[i] != b[i]) return 0*printf("-1
");
    }
    int ans = 0;
    for(int i = 0; i < n; i ++) {
        if(s1[i] != s2[i]) {
            for(j = i; j < n; j ++) {
                if(s1[j] == s2[i])break;
            }
            for(int k = j-1; k >= i; k --) {
                ans++;
                swap(s1[k],s1[k+1]);
            }
        }
    }
    printf("%d
",ans);
    for(int i = 0; i < n; i ++) {
        if(s3[i] != s2[i]) {
            for(j = i; j < n; j ++) {
                if(s3[j] == s2[i])break;
            }
            for(int k = j-1; k >= i; k --) {
                printf("%d ",k+1);
                swap(s3[k],s3[k+1]);
            }
        }
    }
    return 0;
}

C. Songs Compression

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Ivan has $\mathrm{nn songs\; on\; his\; phone.\; The\; size\; of\; the ii-th\; song\; is aiai bytes.\; Ivan\; also\; has\; a\; flash\; drive\; which\; can\; hold\; at\; most mm bytes\; in\; total.\; Initially,\; his\; flash\; drive\; is\; empty.}$

Ivan wants to copy all $\mathrm{nn songs\; to\; the\; flash\; drive.\; He\; can\; compress\; the\; songs.\; If\; he\; compresses\; the ii-th\; song,\; the\; size\; of\; the ii-th\; song\; reduces\; from aiai to bibi bytes\; (bi<aibi<ai).}$

Ivan can compress any subset of the songs (possibly empty) and copy all the songs to his flash drive if the sum of their sizes is at most $\mathrm{mm.\; He\; can\; compress any subset\; of\; the\; songs\; (not\; necessarily\; contiguous).}$

Ivan wants to find the minimum number of songs he needs to compress in such a way that all his songs fit on the drive (i.e. the sum of their sizes is less than or equal to $\mathrm{mm).}$

If it is impossible to copy all the songs (even if Ivan compresses all the songs), print "-1". Otherwise print the minimum number of songs Ivan needs to compress.

Input

The first line of the input contains two integers $\mathrm{nn and mm (1\le n\le 105,1\le m\le 1091\le n\le 105,1\le m\le 109)\; —\; the\; number\; of\; the\; songs\; on\; Ivan\text{'}s\; phone\; and\; the\; capacity\; of\; Ivan\text{'}s\; flash\; drive.}$

The next $\mathrm{nn lines\; contain\; two\; integers\; each:\; the ii-th\; line\; contains\; two\; integers aiai and bibi (1\le ai,bi\le 1091\le ai,bi\le 109, ai>biai>bi)\; —\; the\; initial\; size\; of\; the ii-th\; song\; and\; the\; size\; of\; the ii-th\; song\; after\; compression.}$

Output

If it is impossible to compress a subset of the songs in such a way that all songs fit on the flash drive, print "-1". Otherwise print the minimum number of the songs to compress.

Examples

input

Copy

4 21
10 8
7 4
3 1
5 4

output

Copy

2

input

Copy

4 16
10 8
7 4
3 1
5 4

output

Copy

-1

Note

In the first example Ivan can compress the first and the third songs so after these moves the sum of sizes will be equal to $\mathrm{8+7+1+5=21\le 218+7+1+5=21\le 21.\; Also\; Ivan\; can\; compress\; the\; first\; and\; the\; second\; songs,\; then\; the\; sum\; of\; sizes\; will\; be\; equal 8+4+3+5=20\le 218+4+3+5=20\le 21.\; Note\; that\; compressing\; any\; single\; song\; is\; not\; sufficient\; to\; copy\; all\; the\; songs\; on\; the\; flash\; drive\; (for\; example,\; after\; compressing\; the\; second\; song\; the\; sum\; of\; sizes\; will\; be\; equal\; to 10+4+3+5=22>2110+4+3+5=22>21).}$

In the second example even if Ivan compresses all the songs the sum of sizes will be equal $\mathrm{8+4+1+4=17>168+4+1+4=17>16.}$

贪心问题。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e5+10;
struct Nod{
    ll a, b;
}e[N];
int a[N], b[N];
bool cmp(const Nod &a, const Nod &b) {
    return (a.a-a.b) > (b.a-b.b);
}
int main() {
    ll n, m, sum = 0, ans = 0;
    cin >> n >> m;
    for(int i = 0; i < n; i ++) cin >> e[i].a >> e[i].b, sum += e[i].a;
    sort(e,e+n,cmp);
    for(int i = 0; i < n; i ++) {
        if(sum <= m) break;
        sum -= (e[i].a-e[i].b);
        ans++;
    }
    if(sum > m) printf("-1
");
    else printf("%lld
",ans);
    return 0;
}

D. Walking Between Houses

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are $\mathrm{nn houses\; in\; a\; row.\; They\; are\; numbered\; from 11 to nn in\; order\; from\; left\; to\; right.\; Initially\; you\; are\; in\; the\; house 11.}$

You have to perform $\mathrm{kk moves\; to\; other\; house.\; In\; one\; move\; you\; go\; from\; your\; current\; house\; to\; some\; other\; house.\; You\; can\text{'}t\; stay\; where\; you\; are\; (i.e.,\; in\; each\; move\; the\; new\; house\; differs\; from\; the\; current\; house).\; If\; you\; go\; from\; the\; house xx to\; the\; house yy,\; the\; total\; distance\; you\; walked\; increases\; by |x-y||x-y| units\; of\; distance,\; where |a||a| is\; the\; absolute\; value\; of aa.\; It\; is\; possible\; to\; visit\; the\; same\; house\; multiple\; times\; (but\; you\; can\text{'}t\; visit\; the\; same\; house\; in\; sequence).}$

Your goal is to walk exactly $\mathrm{ss units\; of\; distance\; in\; total.}$

If it is impossible, print "NO". Otherwise print "YES" and any of the ways to do that. Remember that you should do exactly $\mathrm{kk moves.}$

Input

The first line of the input contains three integers $\mathrm{nn, kk, ss (2\le n\le 1092\le n\le 109, 1\le k\le 2\cdot 1051\le k\le 2\cdot 105, 1\le s\le 10181\le s\le 1018)\; —\; the\; number\; of\; houses,\; the\; number\; of\; moves\; and\; the\; total\; distance\; you\; want\; to\; walk.}$

Output

If you cannot perform $\mathrm{kk moves\; with\; total\; walking\; distance\; equal\; to ss,\; print\; "NO".}$

Otherwise print "YES" on the first line and then print exactly $\mathrm{kk integers hihi (1\le hi\le n1\le hi\le n)\; on\; the\; second\; line,\; where hihi is\; the\; house\; you\; visit\; on\; the ii-th\; move.}$

For each $\mathrm{jj from 11 to k-1k-1 the\; following\; condition\; should\; be\; satisfied: hj\ne hj+1hj\ne hj+1.\; Also h1\ne 1h1\ne 1 should\; be\; satisfied.}$

Examples

input

Copy

10 2 15

output

Copy

YES
10 4 

input

Copy

10 9 45

output

Copy

YES
10 1 10 1 2 1 2 1 6 

input

Copy

10 9 81

output

Copy

YES
10 1 10 1 10 1 10 1 10 

input

Copy

10 9 82

output

Copy

NO


这卡了一个多小时，没有考虑k > s 的情况，不然E1 和E2 也可以过的。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e5+10;

int main() {
    ll n, k, s, t;
    cin >> n >> k >> s;
    if((n-1)*k < s || k > s) return 0*printf("NO
");
    ll now = 1, pre = 1, flag = 1;
    t = k;
    printf("YES
");
    for(int i = 1; i <= t; i ++) {
        if(i == t) {
            if(now+s <= n) return 0*printf("%d
",now+s);
            else return 0*printf("%d
",now-s);
        }
        if(s-n+1 >= k-1) {
            if(flag==1)now += n-1, flag = 2;
            else now -= n-1, flag = 1;
            s -= n-1;
        } else {
            if(now == 1) {
                now++;
            } else if(now == 2) now--;
            else if(now == n) now--;
            else if(now == n-1) now++;
            s --;
            // int cnt = s-k+1;
            // if(now-cnt >= 1) now -= cnt;
            // else if(now+cnt <= n) now += cnt;
            // s -= cnt;
        }
        k--;
        printf("%d ",now);
        pre = now;
    }
    return 0;
}

E1. Stars Drawing (Easy Edition)

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A star is a figure of the following type: an asterisk character '*' in the center of the figure and four rays (to the left, right, top, bottom) of the same positive length. The size of a star is the length of its rays. The size of a star must be a positive number (i.e. rays of length $\mathrm{00 are\; not\; allowed).}$

Let's consider empty cells are denoted by '.', then the following figures are stars:

The leftmost figure is a star of size $\mathrm{11,\; the\; middle\; figure\; is\; a star of\; size 22 and\; the\; rightmost\; figure\; is\; a star of\; size 33.}$

You are given a rectangular grid of size $\mathrm{n\times mn\times m consisting\; only\; of\; asterisks\; \text{'}*\text{'}\; and\; periods\; (dots)\; \text{'}.\text{'}.\; Rows\; are\; numbered\; from 11 to nn,\; columns\; are\; numbered\; from 11 to mm.\; Your\; task\; is\; to\; draw\; this\; grid\; using any number\; of stars or\; find\; out\; that\; it\; is\; impossible. Stars can\; intersect,\; overlap\; or\; even\; coincide\; with\; each\; other.\; The\; number\; of stars in\; the\; output\; can\text{'}t\; exceed n\cdot mn\cdot m.\; Each\; star\; should\; be\; completely\; inside\; the\; grid.\; You\; can\; use\; stars\; of\; same\; and\; arbitrary\; sizes.}$

In this problem, you do not need to minimize the number of stars. Just find any way to draw the given grid with at most $\mathrm{n\cdot mn\cdot m stars.}$

Input

The first line of the input contains two integers $\mathrm{nn and mm (3\le n,m\le 1003\le n,m\le 100)\; —\; the\; sizes\; of\; the\; given\; grid.}$

The next $\mathrm{nn lines\; contains mm characters\; each,\; the ii-th\; line\; describes\; the ii-th\; row\; of\; the\; grid.\; It\; is\; guaranteed\; that\; grid\; consists\; of\; characters\; \text{'}*\text{'}\; and\; \text{'}.\text{'}\; only.}$

Output

If it is impossible to draw the given grid using stars only, print "-1".

Otherwise in the first line print one integer $\mathrm{kk (0\le k\le n\cdot m0\le k\le n\cdot m)\; —\; the\; number\; of stars needed\; to\; draw\; the\; given\; grid.\; The\; next kk lines\; should\; contain\; three\; integers\; each\; — xjxj, yjyj and sjsj,\; where xjxj is\; the\; row\; index\; of\; the\; central star character, yjyj is\; the\; column\; index\; of\; the\; central star character\; and sjsj is\; the\; size\; of\; the star.\; Each star should\; be\; completely\; inside\; the\; grid.}$

Examples

input

Copy

6 8
....*...
...**...
..*****.
...**...
....*...
........

output

Copy

3
3 4 1
3 5 2
3 5 1

input

Copy

5 5
.*...
****.
.****
..**.
.....

output

Copy

3
2 2 1
3 3 1
3 4 1

input

Copy

5 5
.*...
***..
.*...
.*...
.....

output

Copy

-1

input

Copy

3 3
*.*
.*.
*.*

output

Copy

-1

Note

In the first example the output

2
3 4 1
3 5 2

is also correct.

E1和E2都是下面这个代码，暴力。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1100;
char s[N][N];
bool vis[N][N];
struct Nod {
    int x, y, s;
}e[N*N];
std::vector<Nod> vs;
int n, m;
void solve(int x, int y) {
    int ans = 0, cnt = 1;
    while(x-cnt >= 1 && x+cnt <= n && y-cnt >= 1 && y+cnt <= m) {
        if(s[x-cnt][y] == '*' && s[x+cnt][y] == '*' && s[x][y+cnt] == '*' && s[x][y-cnt] == '*') {
            ans++;
            vis[x-cnt][y]=vis[x+cnt][y]=vis[x][y-cnt]=vis[x][y+cnt]=vis[x][y] = 1;
        } else break;
        cnt++;
    }
    if(ans == 0) return;
    struct Nod p;
    p.x = x, p.y = y, p.s = ans;
    vs.push_back(p);
}
int main() {
    cin >> n >> m;
    for(int i = 1; i <= n; i ++) cin >> s[i]+1;
    for(int i = 2; i <= n-1; i ++) {
        for(int j = 2; j <= m-1; j ++) {
            if(s[i][j] == '*') solve(i,j);
        }
    }
    for(int i = 1; i <= n; i ++) {
        for(int j = 1; j <= m; j ++) {
            if(s[i][j]=='*') {
                if(vis[i][j]==0) return 0*printf("-1
");
            }
        }
    }
    printf("%d
",vs.size());
    for(int i = 0; i < vs.size(); i ++) {
        printf("%d %d %d
",vs[i].x,vs[i].y,vs[i].s);
    }
    return 0;
}