[弱校联萌2016]2016弱校联盟十一专场10.5

比赛链接：https://www.bnuoj.com/v3/contest_show.php?cid=8506#info

I.裸数位dp，dp[l][pre][dir][fz]表示长度为l的时候上一个数是pre，此时是统计dir（分为增、减、相同），fz记录当前是否是前导零。有种情况，就是各位相同的情况，要减掉。

ps：其实不用fz这一维，删掉就行了，智障了。

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const int maxn = 22;
// l, pre, dir, fz
int dp[maxn][11][3][2];
int digit[maxn];
LL l, r;

LL dfs(int l, int pre, int dir, bool fz, bool flag) {
  if(l == 0) return 1;
  if(!flag && ~dp[l][pre][dir][fz]) return dp[l][pre][dir][fz];
  LL ret = 0;
  int pos = flag ? digit[l] : 9;
  for(int i = 0; i <= pos; i++) {
    if(fz&&(i==0)) ret += dfs(l-1,pre,dir,fz,flag&&(i==pos));
    else {
      if(dir==1) {
        if(i>=pre) ret += dfs(l-1,i,dir,fz&&(i==0),flag&&(i==pos));
      }
      else if(dir == 0) {
        if(i<=pre) ret += dfs(l-1,i,dir,fz&&(i==0),flag&&(i==pos));
      }
      else if(dir == 2) {
        if(i==pre||(pre==10&&i)) ret += dfs(l-1,i,dir,fz&&(i==0),flag&&(i==pos));
      }
    }
  }
  if(!flag) dp[l][pre][dir][fz] = ret;
  return ret;
}

LL f(LL x) {
  int pos = 0;
  while(x) {
    digit[++pos] = x % 10;
    x /= 10;
  }
  return dfs(pos,10,0,true,true)+dfs(pos,0,1,true,true)-dfs(pos,10,2,true,true);
}

inline bool scan_d(LL &num) {
    char in;bool IsN=false;
    in=getchar();
    if(in==EOF) return false;
    while(in!='-'&&(in<'0'||in>'9')) in=getchar();
    if(in=='-'){ IsN=true;num=0;}
    else num=in-'0';
    while(in=getchar(),in>='0'&&in<='9'){
        num*=10,num+=in-'0';
    }
    if(IsN) num=-num;
    return true;
}

int main() {
  // freopen("in", "r", stdin);
  memset(dp, -1, sizeof(dp));
  int T;
  scanf("%d", &T);
  while(T--) {
    scan_d(l); scan_d(r);
    printf("%lld\n", f(r)-f(l-1));
  }
  return 0;
}

F.矩阵快速幂就可以，先找斐波那契数列在%20160516下的循环节，求出fib对应下标再求一遍。参考了这个http://blog.csdn.net/tsoi_vergil/article/details/52718426

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const int maxn = 5;
const LL mod = 20160519;
const LL mod1 = 26880696;
int n;

typedef struct Matrix {
    LL m[maxn][maxn];
    int r;
    int c;
    Matrix(){
        r = c = 0;
        memset(m, 0, sizeof(m));
    }
} Matrix;

Matrix mul(Matrix m1, Matrix m2, LL mod) {
    Matrix ans = Matrix();
    ans.r = m1.r;
    ans.c = m2.c;
    for(int i = 1; i <= m1.r; i++) {
        for(int j = 1; j <= m2.r; j++) {
                 for(int k = 1; k <= m2.c; k++) {
                if(m2.m[j][k] == 0) continue;
                ans.m[i][k] = ((ans.m[i][k] + m1.m[i][j] * m2.m[j][k] % mod) % mod) % mod;
            }
        }
    }
    return ans;
}

Matrix quickmul(Matrix m, int n, LL mod) {
    Matrix ans = Matrix();
    for(int i = 1; i <= m.r; i++) {
        ans.m[i][i]  = 1LL;
    }
    ans.r = m.r;
    ans.c = m.c;
    while(n) {
        if(n & 1) {
            ans = mul(m, ans, mod);
        }
        m = mul(m, m, mod);
        n >>= 1;
    }
    return ans;
}

int main() {
    //freopen("in", "r", stdin);
    int T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d",&n);
        Matrix A; A.c = A.r = 2;
        A.m[1][1] = 1; A.m[1][2] = 1; A.m[2][1] = 1; A.m[2][2] = 0;
        Matrix B; B.c = B.r = 2;
        B = quickmul(A,n,mod1);
        n = B.m[1][2];
        A.m[1][1] = 1; A.m[1][2] = 1; A.m[2][1] = 1; A.m[2][2] = 0;
        B = quickmul(A,n,mod);
        printf("%lld\n",B.m[1][2]%mod);
    }
    return 0;
}

 A.用个线段树维护下就可以，也可以离线+倍增。

#include <bits/stdc++.h>
using namespace std;

#define lrt rt << 1
#define rrt rt << 1 | 1
const int maxn = 100100;
const int maxm = maxn << 2;
char s[maxn];
int q, n;
int asc[256];
typedef struct Pair {
  int x, y;
  Pair(){}
  Pair(int x,int y):x(x),y(y){}
}Pair;
typedef struct Node {
  Pair s[4];
  Node() { for(int i = 0; i < 4; i++) s[i].x = i, s[i].y = 0; }
}Node;

Node sum[maxm];

Node f(Node a, Node b) {
  Node ret;
  for(int i = 0; i < 4; i++) {
      int x = b.s[a.s[i].x].x;
      int y = (a.s[i].x - i + 4) % 4 + (x - a.s[i].x + 4) % 4;
      ret.s[i] = Pair(x, a.s[i].y+b.s[a.s[i].x].y+(y>=4));
    }
  return ret;
}
void build(int rt, int l, int r) {
  sum[rt] = Node();
  if(l == r) {
      sum[rt].s[asc[s[l]]] = Pair((asc[s[l]]+1)%4, 0);
      return;
  }
  int m = (l + r) >> 1;
  build(lrt, l, m); build(rrt, m+1, r);
  sum[rt] = f(sum[lrt], sum[rrt]);
}
Node query(int rt, int l, int r, int L, int R) {
    if(L <= l && r <= R) return sum[rt];
  int m = (l + r) >> 1;
  if(R <= m) return query(lrt,l,m,L,R);
  else if(m < L) return query(rrt,m+1,r,L,R);
  else return f(query(lrt,l,m,L,m), query(rrt,m+1,r,m+1,R));
}

int main() {
  //freopen("in","r",stdin);
  asc['e'] = 0; asc['a'] = 1; asc['s'] = 2; asc['y'] = 3;
  int l, r;
  while(~scanf("%s%d",s+1,&q)) {
      n = strlen(s+1);
      build(1,1,n);
      while(q--) {
        scanf("%d%d",&l,&r);
        if(l > r) swap(l, r);
        if(r-l+1<4) puts("0");
        else printf("%d\n", query(1,1,n,l,r).s[0].y);
      }
  }
  return 0;
}