【BZOJ】4542: [Hnoi2016]大数

题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=4542

$${\sum_{l=1}^{n}\sum_{r=i}^{n}\left [ \sum _{i=l}^{r}S[i]*10^{r-i} \,\,\,\,MOD\,\,\,\,P=0 \right ]}$$

　　注意：$p=2,5$的情况中并不满足以上性质，记一下前缀和然后特判即可。

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<algorithm>
  4 #include<vector>
  5 #include<cstdlib>
  6 #include<cmath>
  7 #include<cstring>
  8 using namespace std;
  9 #define maxn 300100
 10 #define llg long long
 11 #define yyj(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
 12 llg n,m,p,quan[maxn],tail,val[maxn],len,se[maxn],KUAI;
 13 long long ans,Ans[maxn],jisuan1[maxn],jisuan2[maxn];
 14 char s[maxn];
 15 struct Q_
 16 {
 17     llg l,r,num;
 18 }ask[maxn];
 19
 20 bool cmp(const Q_&a,const Q_&b) {if (a.l/KUAI==b.l/KUAI) return a.r<b.r; else return a.l/KUAI<b.l/KUAI; }
 21
 22 llg find(llg x)
 23 {
 24     llg l=1,r=len,wz=-1,mid;
 25     while (l<=r)
 26     {
 27         mid=(l+r)>>1;
 28         if (quan[mid]<=x) {wz=mid; l=mid+1;}else {r=mid-1;}
 29     }
 30     return wz;
 31 }
 32
 33 void solve()
 34 {
 35     cin>>m;
 36     for(int i=1;i<=n;i++)
 37     {
 38         jisuan1[i]=jisuan1[i-1]; jisuan2[i]=jisuan2[i-1];
 39         if( (s[i]-‘0‘)%p==0 )
 40              {
 41             jisuan1[i]++;
 42             jisuan2[i]+=i;
 43              }
 44
 45     }
 46     for(int i=1;i<=m;i++)
 47     {
 48         llg x,y;
 49         scanf("%lld%lld",&x,&y);
 50         printf("%lld\n",jisuan2[y]-jisuan2[x-1] - (x-1)*(jisuan1[y]-jisuan1[x-1]));
 51     }
 52 }
 53
 54 void init()
 55 {
 56     cin>>p;
 57     scanf("%s",s+1);
 58     n=strlen(s+1);
 59     KUAI=sqrt(n)+1;
 60     if (p==5 || p==2) return ;
 61     cin>>m;
 62     for (llg i=1;i<=m;i++) scanf("%lld%lld",&ask[i].l,&ask[i].r),ask[i].r++,ask[i].num=i;
 63     sort(ask+1,ask+m+1,cmp);
 64     llg C=1;
 65     tail=1;quan[1]=0;
 66     for (llg i=n;i>=1;i--)
 67     {
 68         val[i]=val[i+1]+C*(s[i]-‘0‘);
 69         val[i]%=p;
 70         quan[++tail]=val[i];
 71         C*=10; C%=p;
 72     }
 73     sort(quan+1,quan+tail+1);
 74     len=unique(quan+1,quan+tail+1)-(quan+1);
 75     for (llg i=1;i<=n+1;i++) val[i]=find(val[i]);
 76 }
 77
 78 int main()
 79 {
 80     yyj("number");
 81     init();
 82     if (p==2 || p==5) {solve(); return 0;}
 83
 84     llg l=ask[1].l,r=ask[1].r;
 85     for (llg i=l;i<=r;i++)
 86     {
 87         ans+=se[val[i]];
 88         se[val[i]]++;
 89     }
 90     Ans[ask[1].num]=ans;
 91     for (llg k=2;k<=m;k++)
 92     {
 93         llg nl=ask[k].l,nr=ask[k].r;
 94         if (nr>r)
 95         {
 96             for (llg i=r+1;i<=nr;i++)
 97             {
 98                 ans+=se[val[i]];
 99                 se[val[i]]++;
100             }
101         }
102         if (nr<r)
103         {
104             for (llg i=r;i>nr;i--)
105             {
106                 ans-=se[val[i]]-1;
107                 se[val[i]]--;
108             }
109         }
110         if (l<nl)
111         {
112             for (llg i=l;i<nl;i++)
113             {
114                 ans-=se[val[i]]-1;
115                 se[val[i]]--;
116             }
117         }
118         if (l>nl)
119         {
120             for (llg i=l-1;i>=nl;i--)
121             {
122                 ans+=se[val[i]];
123                 se[val[i]]++;
124             }
125         }
126         Ans[ask[k].num]=ans;
127         l=nl,r=nr;
128     }
129     for (llg i=1;i<=m;i++) printf("%lld\n",Ans[i]);
130     return 0;
131 }