#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cmath>
#include<set>
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define enter putchar('\n')
using namespace std; typedef long long ll; const ll p = 1e9 + 7; const int M = 25000005; ll read() { ll ans = 0,op = 1; char ch = getchar(); while(ch < '0' || ch > '9') { if(ch == '-') op = -1; ch = getchar(); } while(ch >= '0' && ch <= '9') { ans *= 10; ans += ch - '0'; ch = getchar(); } return ans * op; } ll qpow(ll a,ll b) { ll q = 1;a %= p; while(b) { if(b & 1) q *= a,q %= p; a *= a,a %= p; b >>= 1; } return q % p; } ll sum[M+1],inv[M+1],n,m,tot,cur,ans,k; char s[5005][5005]; void init() { sum[0] = sum[1] = 1; inv[0] = inv[1] = 1;//inv[i]存i！的逆元 rep(i,2,M-1) sum[i] = sum[i-1] * i % p; inv[M-1] = qpow(sum[M-1],p-2); per(i,M-2,2) inv[i] = inv[i+1] * (i+1) % p; } ll C(ll a,ll b) { return sum[a] * inv[b] % p * inv[a-b] % p; } int main() { init(); while(1) { n = read(),m = read(); if(!n && !m) break; tot = cur = 0; rep(i,1,n) { scanf("%s",s[i]); rep(j,0,m-1) { if(s[i][j] == '0') continue; if(s[i][j] == '+') tot++; if(s[i][j] == '-') tot--; if(s[i][j] == '*') cur++; } } ll ans = 0; if(tot + cur <= 0) printf("0\n"); else if(tot - cur > 0) printf("%lld\n",qpow(2,n*m)); else { rep(i,(cur - tot >> 1) + 1,cur) { ans += C(cur,i); if(ans > p) ans -= p; } ans *= qpow(2,n*m-cur),ans %= p; printf("%lld\n",ans); } } return 0; }

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<iostream>
#include<queue>
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define enter putchar('\n')using namespace std; typedef long long ll; const int M = 1005; typedef pair<ll,ll> pr; ll read() { ll ans = 0,op = 1; char ch = getchar(); while(ch < '0' || ch > '9') { if(ch == '-') op = -1; ch = getchar(); } while(ch >= '0' && ch <= '9') { ans *= 10; ans += ch - '0'; ch = getchar(); } return ans * op; } struct mis { ll a,b; }mi[20]; ll n,m,q,e,g[30][30],s,t,l,d; ll dp[30][700000],st[30]; bool c[30],v[30],vis[30][700000]; queue <pr> pq; void floyd()//floyd处理最短距离 { rep(k,1,n) rep(i,1,n) rep(j,1,n) g[i][j] = min(g[i][j],g[i][k] + g[k][j]); } void build() { memset(g,127/3,sizeof(g)); memset(dp,127/3,sizeof(dp)); n = read(),m = read(),q = read(),e = read(); rep(i,1,n) c[i] = read(),g[i][i] = 0; rep(i,1,n) v[i] = read(); rep(i,1,m) { s = read(),t = read(),l = read(),d = read(); g[s][t] = min(g[s][t],d*l); g[t][s] = min(g[t][s],d*l);//使用邻接矩阵以用floyd  } rep(i,0,q-1) mi[i].a = read(),mi[i].b = read(); rep(i,1,n) { if(!v[i]) { dp[i][0] = 0;//第一维表示枚举的任务，第二维表示状态 pq.push(make_pair(i,0));//如果安全就压入 rep(j,1,n) if(c[j]) g[i][j] = min(g[i][j],e);//建立单向边  } } floyd(); } void cal(ll x) { rep(i,0,q-1) st[i] = x % 3,x /= 3;//将状态转化为三进制串 } ll lac() { ll res = 0; per(i,q-1,0) res *= 3,res += st[i];//三进制串转化回去 return res; } void update(int i,ll val) { st[i]++; ll pos = lac(); st[i]--; ll c; if(st[i] == 0) c = mi[i].a; else c = mi[i].b; dp[c][pos] = min(dp[c][pos],val);//更新距离 if(!vis[c][pos]) vis[c][pos] = 1,pq.push(make_pair(c,pos)); //记录一下此状态已经被走过 } int main() { build(); ll r = 100000000000000; while(!pq.empty()) { pr c = pq.front(); pq.pop(); cal(c.second); bool ret = 1; rep(i,0,q-1) { if(st[i] < 2)//如果当前任务没做完  { ret = 0; if(st[i] == 0) update(i,dp[c.first][c.second] + g[c.first][mi[i].a]); else update(i,dp[c.first][c.second] + g[c.first][mi[i].b]);//更新即可  } } if(ret) r = min(r,dp[c.first][c.second]);//全做完了就更新答案  } printf("%lld\n",r); return 0; } /* 5 4 2 1 1 0 0 0 0 0 1 1 1 0 1 2 1 2 2 3 0 2000 3 4 2 2 4 5 4 2 3 1 4 5 */

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<iostream>
#include<set> #include<vector> #include<utility> #include<map> #define rep(i,a,n) for(int i = a;i <= n;i++) #define per(i,n,a) for(int i = n;i >= a;i--) #define enter putchar('\n') using namespace std; typedef long long ll; const int M = 500005; ll read() { ll ans = 0,op = 1; char ch = getchar(); while(ch < '0' || ch > '9') { if(ch == '-') op = -1; ch = getchar(); } while(ch >= '0' && ch <= '9') { ans *= 10; ans += ch - '0'; ch = getchar(); } return ans * op; } ll n,q,ra[M],xl[M],a,b,x,cnt,keyl[M],keyr[M]; double key[M]; struct node//存储左右端点和下标 { ll dl,dr,id; bool operator < (const node &g) const { return dl < g.dl; } }; struct card//存储卡池的情况，按题目要求重载，之后直接取堆首 { double v; ll pos; bool operator < (const card &g) const { if(v == g.v) return pos < g.pos; return v > g.v; } }; set <node> nq; set <card> cq; struct seg { ll sum[M<<2],exp[M<<2]; void update(ll p) { sum[p] = sum[p<<1] + sum[p<<1|1]; exp[p] = exp[p<<1] + exp[p<<1|1]; } void build(ll p,ll l,ll r,ll num) { if(l == r) //建树的时候要对应稀有度  { if(num == ra[l]) sum[p] = 1,exp[p] = xl[l]; return; } ll mid = l+r >> 1; build(p<<1,l,mid,num); build(p<<1|1,mid+1,r,num); update(p); } void modify(ll p,ll l,ll r,ll pos,ll val)//常规修改  { if(l == r) { exp[p] = val; return; } ll mid = l+r >> 1; if(pos <= mid) modify(p<<1,l,mid,pos,val); else modify(p<<1|1,mid+1,r,pos,val); exp[p] = exp[p<<1] + exp[p<<1|1]; } double query1(ll p,ll l,ll r,ll kl,ll kr)//找喜爱度  { if(l == kl && r == kr) return (double)exp[p]; ll mid = l + r >> 1; if(kr <= mid) return query1(p<<1,l,mid,kl,kr); else if(kl > mid) return query1(p<<1|1,mid+1,r,kl,kr); else return query1(p<<1,l,mid,kl,mid) + query1(p<<1|1,mid+1,r,mid+1,kr); } double query2(ll p,ll l,ll r,ll kl,ll kr)//找个数  { if(l == kl && r == kr) return (double)sum[p]; ll mid = l + r >> 1; if(kr <= mid) return query2(p<<1,l,mid,kl,kr); else if(kl > mid) return query2(p<<1|1,mid+1,r,kl,kr); else return query2(p<<1,l,mid,kl,mid) + query2(p<<1|1,mid+1,r,mid+1,kr); } }t[4]; double calc(ll l,ll r)//丧心病狂的计算（注意这里是从0开始的否则会出错） { double ans = 0; ans += 10.0 * (9.0 / 10.0 * t[0].query1(1,1,n,l,r) / t[0].query2(1,1,n,l,r)); ans += 10.0 * (9.0 / 100.0 * t[1].query1(1,1,n,l,r) / t[1].query2(1,1,n,l,r)); ans += 10.0 * (t[2].query1(1,1,n,l,r) / t[2].query2(1,1,n,l,r) / 100.0); ans += 9.0 / 10.0 * t[1].query1(1,1,n,l,r) / t[1].query2(1,1,n,l,r); ans += t[2].query1(1,1,n,l,r) / t[2].query2(1,1,n,l,r) / 10.0; return ans; } int main() { n = read(),q = read(); rep(i,1,n) ra[i] = read(); rep(i,1,n) xl[i] = read(); rep(i,0,2) t[i].build(1,1,n,i); key[0] = calc(1,n),keyl[0] = 1,keyr[0] = n;//记录下标对应的值，左右端点的位置 // printf("%.4lf\n",key[0]); cq.insert((card){key[0],0}); nq.insert((node){1,n,0}); rep(i,1,q) { set <card> :: iterator it1; // for(it1 = cq.begin();it1 != cq.end();it1++) printf("%.4lf %lld\n",it1->v,it1->pos); char c = getchar(); if(c == '1')//使用字符读入处理……当然有其他方式  { a = read(),b = read(); key[++cnt] = calc(a,b),keyl[cnt] = a,keyr[cnt] = b;//新建卡池并插入堆中  cq.insert((card){key[cnt],cnt}); nq.insert((node){a,b,cnt}); } else if(c == '2')//把卡池删了  { x = read(); cq.erase((card){key[x],x}); nq.erase((node){keyl[x],keyr[x],x}); } else if(c == '3') { a = read(),b = read(); t[ra[a]].modify(1,1,n,a,b); cq.erase((card){key[0],0}); key[0] = calc(1,n); cq.insert((card){key[0],0});//先改基础卡池 set <node> :: iterator it = nq.upper_bound((node){a,233,233});//找到要改的 // printf("#%lld\n",(*it).id); if(it == nq.begin()) continue;//如果是基础卡池就不改 it--; ll k = (*it).id; // printf("#%lld\n",k); // if((*it).dr < x) continue;//这里很奇怪，不删70，删了AC  cq.erase((card){key[k],k}); key[k] = calc(keyl[k],keyr[k]); cq.insert((card){key[k],k});//删除，修改，重新插入  } else { card now = *(cq.begin()); printf("%lld\n",now.pos);//否则直接输出堆首元素，下面是处理K以外的字符的 c = getchar(),c = getchar(),c = getchar(); } } return 0; }