Bestcoder Round #73 (div.2)

http://acm.hdu.edu.cn/showproblem.php?pid=5630
题意：
黑白相间棋盘，每次可以对一个矩形的格子进行取反，问想要达到颜色都一样最少几次？
思路：
首先，如果先对偶数行取反，再对偶数列取反，可以得到一个[n/2]+[m/2][n/2] + [m/2][n/2]+m/2的解, 只要说明这个这是答案的下界就可以了。 考虑第一列，每次操作最多使得两个第一列的相邻元素变得一样， 第一列有n−1n-1n−1对相邻元素,这样使得第一列变成一样的次数就是[(n−1)/2][(n-1)/2](n−1)/2，同理考虑第一行即可!

#include<bits/stdc++.h>using namespace std;int main(){    int n,m,t;cin>>t;    while(t--){        scanf("%d%d",&n,&m);        printf("%d\n",n/2+m/2);    }}

http://acm.hdu.edu.cn/showproblem.php?pid=5631
题意：
n个点，n+1条边，问有多少种情况使得至少删掉1条边后，整个图仍然联通。
思路：
让 nnn 个点联通最少需要 n−1n-1n−1 条边，所以最多只能删除两条边，我们可以枚举删除的这两条边（或者唯一的一条边），然后暴力BFS判断连通性就好了。时间复杂度 O(n^3)！

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<iostream>using namespace std;int T,n,ans;int u[105],v[105],b[105],f[105];int getf(int x){return f[x]?f[x]=getf(f[x]):x;}int merge(int x,int y){    x=getf(x),y=getf(y);    if(x!=y)return f[x]=y,1;    return 0;}int check(){    memset(f,0,n+1<<2);    int pn=n;    for(int i=0;i<=n;i++)        if(!b[i])            pn-=merge(u[i],v[i]);    return pn==1;}int main(){    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        ans=0;        for(int i=0;i<=n;i++)            scanf("%d%d",u+i,v+i);        for(int i=0;i<=n;i++)        {            b[i]=1;            ans+=check();            for(int j=i+1;j<=n;j++)            {                b[j]=1;                ans+=check();                b[j]=0;            }            b[i]=0;        }        cout<<ans<<endl;    }    return 0;}

http://acm.hdu.edu.cn/showproblem.php?pid=5632

很明显这是一道数位DP的题目，状态 dp[i][j][k]dp[i][j][k]dp[i][j][k] 表示当前考虑了最高的 iii 位，两个数目前数位和的差是 jjj，当前两个数以及给定的 nnn 之间的大小关系是 kkk，然后暴力枚举这两个数的当前位的值，然后转移就好了。时间复杂度 O(log^2 n)

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<map>#include<queue>#include<set>#define mo 998244353#define maxn 1010 #define fr first#define sc second#define mp make_pair#define pb push_back#define son e[i].t#define pow POW#define clear(x) memset(x, 0, sizeof(x))typedef long long ll;using namespace std;int T, n;int cnt[maxn][maxn], sum[maxn][maxn], f[maxn];int o[maxn], a[maxn], len, tmp;char s[maxn];bool divide(){    if (o[1] % 2) a[len] = 1, o[1] --; ++ len;    for (int i = tmp; i >= 1; -- i){        if (o[i] % 2 == 0)            o[i] /= 2;         else{            o[i - 1] += 10;            o[i] = (o[i] - 1) / 2;        }    }    while (tmp && ! o[tmp]) -- tmp;    return tmp > 0;}int now[maxn];void solve(){    clear(a);    scanf("%s", s + 1); tmp = strlen(s + 1); len = 0;    for (int i = 1; i <= tmp; ++ i) o[i] = s[tmp - i + 1] - '0';    o[1] = o[1] + 1;    while (divide());    clear(now);    int tot = 0, ans = 0;    for (int i = len; i >= 0; -- i){        if (a[i]){            for (int j = tot + 1; j <= 1000; ++ j)                 ans = (ans + 1ll * now[j] * sum[i][j - tot - 1] % mo) % mo;            for (int j = tot; j <= 1000; ++ j)                now[j] = (now[j] + cnt[i][j - tot]) % mo;            ++ tot;            ans = (ans + f[i]) % mo;        }    }    printf("%d\n", ans);}int main(){    cnt[0][0] = 1;    for (int i = 0; i <= 1000; ++ i) sum[0][i] = 1;    f[1] = 0;    for (int i = 1; i <= 1000; ++ i){        cnt[i][0] = cnt[i - 1][0];        for (int j = 1; j <= 1000; ++ j)            cnt[i][j] = (cnt[i - 1][j - 1] + cnt[i - 1][j]) % mo;        sum[i][0] = cnt[i][0];        for (int j = 1; j <= 1000; ++ j) sum[i][j] = (sum[i][j - 1] + cnt[i][j]) % mo;        f[i + 1] = 2ll * f[i] % mo;        for (int j = 2; j <= 1000; ++ j) f[i + 1] = (f[i + 1] + 1ll * cnt[i][j] * sum[i][j - 2] % mo) % mo;    }    scanf("%d", &T);    while (T --) solve();}

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <string>#include <vector>#include <deque>#include <stack>#include <queue>#include <algorithm>using namespace std;const int maxn = 1100;const int modu = 998244353;string op , ne;int a[maxn] , cnt;int id(char c) { return c-'0'; }char reid(int c) { return c+'0'; }void turn(){    cnt = 0;    while(true)    {        if(op=="1")        {            a[++cnt] = id(op[0]);            break;        }        if(id(op[op.size()-1])%2) a[++cnt] = 1; else a[++cnt] = 0;        int last = 0 , now;        for(int i=0;i<op.size();i++)        {                now = last*10+id(op[i]);            if(now%2) last = 1; else last = 0;            if(i || now/2) ne += reid(now/2);        }        op = ne; ne="";    }    reverse(a+1, a+1+cnt);}int d[maxn][maxn*2][2][2];void up(int& a , int b) { a = (a+b)%modu; }int main(int argc, char *argv[]) {    int t , n;    cin>>t;    while(t--)    {        cin>>op;        turn();        n = cnt;        for(int i=0;i<=n;i++) for(int j=0;j<=2*n;j++) for(int k=0;k<2;k++) for(int l=0;l<2;l++)             d[i][j][k][l] = 0;        d[0][n][1][1] = 1;        for(int i=0;i<n;i++) for(int j=0;j<=2*n;j++) for(int k=0;k<2;k++) for(int l=0;l<2;l++)        {            int& now = d[i][j][k][l];            if(!now) continue;            int jj , kk , ll;            for(int x=0;x<2;x++) for(int y=0;y<2;y++)            {                if(k && x>a[i+1]) continue;                if(l && x<y) continue;                jj = j+x-y;                if(k && x==a[i+1]) kk = 1; else kk = 0;                if(l && x==y) ll = 1; else ll = 0;                up(d[i+1][jj][kk][ll], now);            }        }        int res = 0;        for(int i=n-1;i>=0;i--) for(int k=0;k<2;k++) up(res, d[n][i][k][0]);        cout<<res<<endl;    }    return 0;}

http://acm.hdu.edu.cn/showproblem.php?pid=5634
注意到10^7​​ 之内的数最多phi O(log(n))O(log(n)) 次就会变成11， 因此可以考虑把一段相同的不为11的数缩成一个点，用平衡树来维护。每次求phi的时候就在平衡树上取出这个区间然后暴力求phi，如果一段数变成了11，就在平衡树里面删掉它，最后统计答案的时候只要把区间中被删去的11加回答案即可，时间复杂度O((n + m)logn)

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const int N = 1e7, L = 7e5, T = 3e5 + 10;bool used[N + 10];int phi[N + 10], p[L];int a[T], flag[T << 2];long long sum[T << 2];int len, t, n, m;void getprime(){    phi[1] = 1;    for(int i = 2; i <= N; ++i)    {        if(! used[i])        {            p[++len] = i;            phi[i] = i - 1;        }        for(int j = 1; j <= len && i * p[j] <= N; ++j)        {            used[i * p[j]] = 1;            if(i % p[j] == 0)            {                phi[i * p[j]] = phi[i] * p[j];                break;            }            else                phi[i * p[j]] = phi[i] * (p[j] - 1);        }    }}void build(int x, int L, int R){    if(L == R)    {        sum[x] = flag[x] = a[L];        return;    }    int mid = (L + R) >> 1;    build(x << 1, L, mid);    build(x << 1 | 1, mid + 1, R);    sum[x] = sum[x << 1] + sum[x << 1 | 1];    flag[x] = (flag[x << 1] == flag[x << 1 | 1] ? flag[x << 1] : 0);}void update2(int x, int L, int R, int tl, int tr, int y){    if(L <= tl && tr <= R)    {        flag[x] = y;        sum[x] = (long long)flag[x] * (tr - tl + 1);        return;    }    int mid = (tl + tr) >> 1;    if(flag[x])    {        update2(x << 1, tl, mid, tl, mid, flag[x]);        update2(x << 1 | 1, mid + 1, tr, mid + 1, tr, flag[x]);    }    if(L <= mid)        update2(x << 1, L, R, tl, mid, y);    if(mid < R)        update2(x << 1 | 1, L, R, mid + 1, tr, y);    sum[x] = sum[x << 1] + sum[x << 1 | 1];    flag[x] = (flag[x << 1] == flag[x << 1 | 1] ? flag[x << 1] : 0);}void update1(int x, int L, int R, int tl, int tr){    if(flag[x] && L <= tl && tr <= R)    {        flag[x] = phi[flag[x]];        sum[x] = (long long)flag[x] * (tr - tl + 1);        return;    }    int mid = (tl + tr) >> 1;    if(flag[x])    {        update2(x << 1, tl, mid, tl, mid, flag[x]);        update2(x << 1 | 1, mid + 1, tr, mid + 1, tr, flag[x]);    }    if(L <= mid)        update1(x << 1, L, R, tl, mid);    if(mid < R)        update1(x << 1 | 1, L, R, mid + 1, tr);    sum[x] = sum[x << 1] + sum[x << 1 | 1];    flag[x] = (flag[x << 1] == flag[x << 1 | 1] ? flag[x << 1] : 0);}long long query(int x, int L, int R, int tl, int tr){    if(L <= tl && tr <= R)        return sum[x];    int mid = (tl + tr) >> 1;    if(flag[x])    {        update2(x << 1, tl, mid, tl, mid, flag[x]);        update2(x << 1 | 1, mid + 1, tr, mid + 1, tr, flag[x]);    }    long long re = 0;    if(L <= mid)        re += query(x << 1, L, R, tl, mid);    if(mid < R)        re += query(x << 1 | 1, L, R, mid + 1, tr);    return re;}int main(){    getprime();    scanf("%d", &t);    while(t--)    {        scanf("%d%d", &n, &m);        for(int i = 1; i <= n; ++i)            scanf("%d", &a[i]);        build(1, 1, n);        int op, x, y, z;        while(m--)        {            scanf("%d%d%d", &op, &x, &y);            if(op == 1)                update1(1, x, y, 1, n);            else if(op == 2)            {                scanf("%d", &z);                update2(1, x, y, 1, n, z);            }            else                printf("%I64d\n", query(1, x, y, 1, n));        }    }    return 0;}