(UVA)11916 Emoogle Grid

Posted 2020-08-04 SJY

题意：一个N列的网格，有B个格子可以不涂色，其他格子各涂一种颜色，现在一共有k种颜色，要求同一列格子颜色不能相同，问总方案数 MOD 100000007答案等于R时最小的M是多少。

思路：首先这个m一定是大于等于所给的不能放的点的x的最大值。我们可以先统计出当前矩阵的方案数，我们知道如果当前格子上面能放东西，当前格子的方案数是k-1,否则就是k种，然后乘法原理，所以先判断下当前的m是否符合，然后不符合的情况下，再加一层，再判断下，算出当前答案cnt，你再往下加层数的时候情况是一样的，都是cnt*(（k-1）^n);然后就转换成

求cnt*(（k-1）^(n*t))%mod = r;然后大步小步算出t的最小值再加上原来m就是答案，复杂度（sqrt(mod)）；

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<stdlib.h>
  4 #include<iostream>
  5 #include<queue>
  6 #include<math.h>
  7 #include<map>
  8 #include<vector>
  9 #include<string.h>
 10 #include<set>
 11 typedef long long LL;
 12 using namespace std;
 13 typedef struct pp {
 14         int x;
 15         int y;
 16 } ss;
 17 ss ans[100005];
 18 ss bns[100005];
 19 set<pair<int,int> >vec;
 20 const  LL mod = 100000007;
 21 bool cmp(pp p, pp q) {
 22         if(p.x == q .x)
 23                 return p.y < q.y;
 24         return p.x < q.x;
 25 }
 26 LL coutt(int m,int n,int b,int k);
 27 LL quick(LL n, LL m);
 28 LL solve(int n,int m,int b,int k,int r);
 29 LL BSS(LL kk,LL k,LL cnt,LL n);
 30 int er(int n,int m,LL fin);
 31 ss cn[100005];
 32 int main(void) {
 33         int T;
 34         int __ca = 0;
 35         scanf("%d",&T);
 36         while(T--) {
 37                 __ca++;
 38                 int  n, k, b,r;
 39                 vec.clear();
 40                 scanf("%d %d %d %d",&n,&k,&b,&r);
 41                 int i,j;
 42                 int  m = 1;
 43                 for(i = 0 ; i  < b; i++) {
 44                         scanf("%d %d",&ans[i].x,&ans[i].y);
 45                         if(ans[i].x > m)m = ans[i].x;
 46                         pair<int,int>ac = make_pair(ans[i].x,ans[i].y);
 47                         // printf("%d %d\n",ans[i].x,ans[i].y);
 48                         vec.insert(ac);
 49                 }
 50                 printf("Case %d: %lld\n",__ca,solve(n,m,b,k,r));
 51         }
 52         return 0;
 53 }
 54 LL coutt(int m,int n,int b,int k) {
 55         LL sum = 0;
 56         for(int i = 0; i < b; i++) {//printf("%d\n",ans[i].x);
 57                 if(ans[i].x != m&& !vec.count(make_pair(ans[i].x+1,ans[i].y)))
 58                         sum++;
 59                 if(ans[i].x == 1) {
 60                         sum--;
 61                 }
 62         }
 63         sum += n;
 64         LL dt = quick((LL)k,sum)%mod;
 65         dt =  dt*quick((LL)(k-1),(LL)m*(LL)n-b-sum)%mod;
 66         return dt % mod;
 67 }
 68 LL quick(LL n, LL m) {
 69         LL ask = 1;
 70         n %= mod;
 71         while(m) {
 72                 if(m&1)
 73                         ask =  ask*n %mod;
 74                 n = n*n %mod;
 75                 m>>=1;
 76         }
 77         return ask ;
 78 }
 79 LL solve(int n,int m,int b,int k,int r) {
 80         LL cnt = coutt(m,n,b,k);
 81         if(cnt == r)
 82                 return (LL)m;
 83         int i,j;
 84         int ac = 0;
 85         for(i = 0; i < b; i++) {
 86                 if(ans[i].x == m) {
 87                         ac++ ;
 88                 }
 89         }
 90         //printf("%lld\n",cnt);
 91         m++;
 92         LL ak = quick((LL)k,ac);
 93         cnt = cnt *ak %mod;
 94         ak = quick((LL)(k-1),n-ac);
 95         cnt = cnt *ak%mod;
 96         if(cnt == r)
 97                 return (LL)m;
 98         LL tp = quick((LL)(k-1),n);
 99         LL kk = (LL)r;
100         return (m+BSS(kk,(LL)k,cnt,n))%mod;
101 }
102 LL BSS(LL kk,LL k,LL cnt,LL n) {
103         int i,j;
104         LL ni = quick(cnt,mod-2);
105         kk = kk*ni%mod;
106         LL ak = quick(k-1,n)%mod;
107         LL modd = sqrt(1.0*mod);
108         LL ac = kk;
109         LL nii = quick(ak,mod-2);
110         for(i = 0; i < modd; i++) {
111                 bns[i].x = ac;
112                 bns[i].y = i;
113                 ac = ac*nii%mod;
114         }
115 
116         sort(bns,bns+modd,cmp);
117         int cnn = 1;
118         cn[0] = bns[0];
119         int c = bns[0].x;
120         for(i = 1; i <modd; i++) {
121                 if(c!=bns[i].x) {
122                         c=bns[i].x;
123                         cn[cnn]=bns[i];
124                         cnn++;
125                 }
126         }
127         LL akk = 1;
128         LL app = quick(ak,modd);
129         for(i = 0; i <= modd+1; i++) {
130                 int ack = er(0,cnn,akk);
131                 if(ack!=-1) {   //printf("%d %lld\n",i,ack);
132                         return modd*(LL)i+(LL)ack;
133                 }
134                 akk = akk*app%mod;
135         }
136 }
137 int er(int n,int m,LL fin) {
138         if(n>m)
139                 return -1;
140         int mid = (n+m)/2;
141         if(cn[mid].x == fin) {
142                 return cn[mid].y;
143         } else if(cn[mid].x > fin) {
144                 return er(n,mid-1,fin);
145         } else return er(mid+1,m,fin);
146 }