【CF521C】Pluses everywhere(贡献)

题意:有一个长为n的数字字符串,要求其中插入k个加号,求所有合法表达式的和之和

0<=k<n<=1e5

思路:参考官方题解,讲的很好很清楚

字符串下标从0开始

考虑第i位d[i]的贡献,分两类讨论

1.d[i]不是最后一个串

枚举i到该串最后一个字符的距离l

d[i]之前未知,从d[i]到d[i+l]之间确定没有加号,d[i+l]后面是加号

有n-1个间隔,确定l+1个没有加号,剩余n-l-2个间隔内要放k-1个加号,方案数即C(n-l-2,k-1)

2.d[i]是最后一个串

d[i]之前未知,从d[i]到d[i+l]之间确定没有加号

有n-1个间隔,确定l个没有加号,剩余n-l-1个间隔内要放k个加号,方案数即C(n-l-1,k)=C(i,k)

最后需要交换一下求和顺序

 

 

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 typedef unsigned int uint;
 5 typedef unsigned long long ull;
 6 typedef long double ld;
 7 typedef pair<int,int> PII;
 8 typedef pair<ll,ll> Pll;
 9 typedef vector<int> VI;
10 typedef vector<PII> VII;
11 typedef pair<ll,ll>P;
12 #define N  200000+10
13 #define M  200000+10
14 #define INF 1e9
15 #define fi first
16 #define se second
17 #define MP make_pair
18 #define pb push_back
19 #define pi acos(-1)
20 #define mem(a,b) memset(a,b,sizeof(a))
21 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
22 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
23 #define lowbit(x) x&(-x)
24 #define Rand (rand()*(1<<16)+rand())
25 #define id(x) ((x)<=B?(x):m-n/(x)+1)
26 #define ls p<<1
27 #define rs p<<1|1
28 #define fors(i) for(auto i:e[x]) if(i!=p)
29 
30 const int MOD=1000000007,inv2=(MOD+1)/2;
31       //int p=1e4+7;
32       double eps=1e-6;
33       int dx[4]={-1,1,0,0};
34       int dy[4]={0,0,-1,1};
35 
36 char ch[N];
37 ll fac[N],inv[N],ten[N],s[N],d[N];
38 
39 int read()
40 {
41    int v=0,f=1;
42    char c=getchar();
43    while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
44    while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
45    return v*f;
46 }
47 
48 ll readll()
49 {
50    ll v=0,f=1;
51    char c=getchar();
52    while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
53    while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
54    return v*f;
55 }
56 
57 ll C(int x,int y)
58 {
59     if(y>x) return 0;
60     return fac[x]*inv[y]%MOD*inv[x-y]%MOD;
61 }
62 
63 int main()
64 {
65     //freopen("1.in","r",stdin);
66     //freopen("1.out","w",stdout);
67     int n=read(),K=read();
68     scanf("%s",ch);
69     rep(i,0,n-1) d[i]=ch[i]-'0';
70     fac[0]=1;
71     rep(i,1,2e5) fac[i]=fac[i-1]*i%MOD;
72     inv[0]=inv[1]=1;
73     rep(i,2,2e5) inv[i]=inv[MOD%i]*(MOD-MOD/i)%MOD;
74     rep(i,1,2e5) inv[i]=inv[i-1]*inv[i]%MOD;
75     ten[0]=1;
76     rep(i,1,2e5) ten[i]=ten[i-1]*10%MOD;
77     s[0]=d[0];
78     rep(i,1,n-1) s[i]=s[i-1]+d[i];
79     ll ans=0;
80     rep(i,0,n-2) ans=(ans+ten[i]*C(n-i-2,K-1)%MOD*s[n-i-2]%MOD)%MOD;
81     rep(i,0,n-1) ans=(ans+d[i]*ten[n-i-1]%MOD*C(i,K)%MOD)%MOD;
82     printf("%I64d\n",ans);
83     return 0;
84 }

 

posted on 2019-11-23 19:17  myx12345  阅读(162)  评论(0编辑  收藏  举报

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