Codeforces Round #423Div. 2 A-E题解

来源：互联网发布：u盘安装linux卡在scsi 编辑：程序博客网时间：2024/05/17 21:57

A题：水题看题目然后模拟一下就好了

#include <bits/stdc++.h>using namespace std;int main(){    int n,o,t;    int tmp=0;    int ans=0;    scanf("%d%d%d",&n,&o,&t);    while(n--)    {        int x;        scanf("%d",&x);        if(x==2)        {            if(t)t--;            else ans+=2;        }        else        {            if(o)o--;            else            {                if(t)                {                    t--;                    tmp++;                }                else                {                    if(tmp)tmp--;                    else ans++;                }            }        }    }    printf("%d\n",ans);}

B题：求长和宽那个更大然后判断这个方阵容不容的下即可

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>using namespace std;char map[105][105];int l,r,u,d;int main(){    int n,m;    while(scanf("%d%d",&n,&m)!=EOF){        l = u = max(n,m);        r = d = -1;        for(int i = 0; i < n; i++)            scanf("%s",map[i]);        int ans = 0;        for(int i = 0; i < n; i++)            for(int j = 0; j < m; j++)                if(map[i][j] == 'B'){                    l = min(l,j),r = max(r,j),u = min(u,i),d = max(d,i);                    ans++;                }        if(r == -1){            puts("1");            continue;        }        int len = max(r-l+1,d-u+1);    //  printf("%d\n",len);         if(n>=len && m >= len){            printf("%d\n",len*len-ans);        }               else             puts("-1");    }    return 0;}

题解思路：路径压缩去找最右边没有被使用过的字符赋上字符即可

#include<iostream>#include<cstring>#include<algorithm>using namespace std;const int mx = 3e6+5;int p[mx];char s[mx];char str[mx];int find(int x){    return p[x] == x?x:p[x] = find(p[x]);}int main(){    int n,m,pos;    for(int i = 0; i < mx; i++)        p[i] = i;    memset(s,0,sizeof(s));    int MAXN = 0;    scanf("%d",&n);    while(n--){        scanf("%s",str);        scanf("%d",&m);        int len = strlen(str);        while(m--){            scanf("%d",&pos);            pos--;            MAXN = max(MAXN,pos+len);            for(int i = find(pos); i-pos<len; i = p[i] = find(i+1))                s[i] = str[i-pos];        }    }    for(int i = 0; i < MAXN; i++)        if(s[i] == 0)            putchar('a');        else            printf("%c",s[i]);    printf("\n");    return 0;}

题解思路：最下面的一个1结着一个然后最后把所有的没有入度为0的点（除n外）都接到第n个接点

然后边数就是如果n%k==0 || n%k==1时n/k×2当n%k == 2时就是n/k×2+1当n%k == 0&&k ==2时就是 n/k×2-1

#include<iostream>#include<cstring>#include<algorithm>#include<cstdio>using namespace std;const int mx = 2e5+5;bool vis[mx];int main(){    int n,k;    while(scanf("%d%d",&n,&k)!=EOF){        int ans = 0;        memset(vis,0,sizeof(vis));        if(n%k==2)            ans = 1;        else if(n%k != 1 && n%k != 0)            ans = 2;            if(n%k == 0 && k == 2)            ans--;        int d = n/k;        ans = ans + 2*d;        printf("%d\n",ans);        for(int i = 0; i < d; i++)            for(int j = 1; j <= k&&(i+1)*k+j<=n; j++){                vis[i*k+j] = 1;                printf("%d %d\n",i*k+j,(i+1)*k+j);            }        for(int i = 1; i < n; i++)            if(!vis[i])                printf("%d %d\n",i,n);    }    return 0;}

题解思路：因为e最长10可以把这些重叠起来叠成每1个一叠，2个。。。10，然后用树状数组维护

求第一叠到第N叠一共有几个A字符，T字符，C字符，G字符就可以了

#include<iostream> #include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int mx = 1e5+5;char s[mx],e[20];int    sum[4][11][11][mx];int mp[256];int MAXN;void add(int k,int n,int p,int m){    for(;m <= MAXN; m += m&(-m)){        sum[k][n][p][m]++;    }}void dec(int k,int n,int p,int m){    for(;m <= MAXN; m += m&(-m))        sum[k][n][p][m]--;}int query(int k,int n,int p,int m){    int ans = 0;    while(m){        ans += sum[k][n][p][m];        m -= m&(-m);    }    return ans;}int main(){    mp['A'] = 0;    mp['C'] = 1;    mp['G'] = 2;    mp['T'] = 3;    int n,m;    scanf("%s",s+1);    int len = strlen(s+1);    for(int i = 1; i <= len; i++)        for(int j = 1; j <= 10; j++){            MAXN = len/j+1;            add(mp[s[i]],j,i%j,i/j+1);        }    scanf("%d",&n);    while(n--){        int casei ;        scanf("%d",&casei);        if(casei == 1){            int p;            char c;            scanf("%d %c",&p,&c);            for(int i = 1; i <= 10; i++){                MAXN = len/i+1;                add(mp[c],i,p%i,p/i+1);                dec(mp[s[p]],i,p%i,p/i+1);            }            s[p] = c;        }        else{             int l,r;             scanf("%d%d%s",&l,&r,e+1);             int len = strlen(e+1);             int ans = 0;             int p1,p2;             int s = (r-l)/len;             for(int i = 1; i <= len; i++){                p1 = l+i-1;                p2 = p1+s*len;                if(p2>r)                    p2-=len;                ans += query(mp[e[i]],len,p2%len,p2/len+1)-query(mp[e[i]],len,p1%len,p1/len);             }             printf("%d\n",ans);        }    }    return 0;}

阅读全文

0 0