题意：给你n个东西，叫你把n分成任意段，这样的分法有几种（例如3:1 1 1，1 2，2 1，3 ；所以3共有4种），n最多有1e5位，答案取模p = 1e9+7

思路：就是往n个东西中间插任意个板子，所以最多能插n - 1个，所以答案为2^(n - 1) % p。直接套用模板

 

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #define ll long long#define N 1000100using namespace std;char b[N];ll p[N];ll a, c;ll quick(ll a, ll b){ //快速幂    ll k = 1;    while(b){        if(b%2==1){            k = k*a;            k %=c;        }        a = a*a%c;        b /=2;    }    return k;}void priem(){    memset(p, 0, sizeof(p));    ll i, j;    p[1] = 1;    for(i=2; i<=sqrt(N); i++){        for(j=2; j<=N/i; j++)            p[i*j] = 1;    }}ll ola(ll n){ //欧拉函数    ll i, j, r, aa;    r = n;    aa = n;    for(i=2; i<=sqrt(n); i++){        if(!p[i]){            if(aa%i==0){                r = r/i*(i-1);                while(aa%i==0)                    aa /= i;            }        }    }    if(aa>1)        r = r/aa*(aa-1);    return r;}int main(){    ll d, i, j;    priem();    a=2;c=1e9+7;    int T;    cin>>T;    while(T--){        scanf("%s",b);        b[strlen(b)-1]--;        ll l = strlen(b);        ll B=0;        ll oc = ola(c);      //  cout<<"oc = "<
          
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            oc) break; } //cout<
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