"Mathematics" CSU 1810 Reverse (2016 Hunan province 12th session of computer Program design Competition)

Topic Links:

http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1810

Main topic:

A decimal number of length n, R (I,J) indicates that the number after flipping the position I to the first J is the MoD 109+7

Topic Ideas:

Mathematical

This is a different way of thinking, to see how each number can contribute to the answer. (Can push 01,10,001,010,100 ...) can also write a violent like me to play a table within 10 to see the law)

Now consider the case where the number of position I is 1 (this is the last number on the line). It can be found that the contribution is symmetrical (1 and 1 of the n-i+1 are the same as the number of times in the other position, so only consider the I<=N/2 case)

The 1 of the bit I is left in the case of I, there are three, I-left interval exchange, I-right interval exchange, with I-centered interval exchange. The number of exchanges can be obtained through O (1).

And I can exchange to other positions of the answer is also good to push, exchange to the first only interval [1,i], swap to the second bit has the interval [1,i+1],[2,i], third ...

It can be found that the number of exchanges to the first I is,... i-1,i,i,..., i,i,i-1,..., 2,1 (the latter half and the first half are symmetrical) (from 1 to i,i flat, from I to 1)

The effect of the ascending sequence of the preceding and subsequent sequences on the final answer can be computed by preprocessing (the following is: 1+2*10+3*100 ... before, remember to take a power of 10), and the middle level can be prefixed and calculated (10 power interval and *i)

So this problem can be done. Enumerate each bit of I and add its contribution to the answer as above.

1 //2 //by Coolxxx3 //#include <bits/stdc++.h>4#include <iostream>5#include <algorithm>6#include <string>7#include <iomanip>8#include <map>9#include <stack>Ten#include <queue> One#include <Set> A#include <bitset> -#include <memory.h> -#include <time.h> the#include <stdio.h> -#include <stdlib.h> -#include <string.h> - //#include <stdbool.h> +#include <math.h> - #defineMin (a) < (b) ( A):(B)) + #defineMax (a) (a) > (b)? ( A):(B)) A #defineABS (a) ((a) >0? ( A):(-(a))) at #defineLowbit (a) (a& (a)) - #defineSqr (a) ((a) * (a)) - #defineSwap (a) (a) ^= (b), (b) ^= (a), (a) ^= (b)) - #defineMem (A, B) memset (A,b,sizeof (a)) - #defineEPS (1E-8) - #defineJ 10000 in #defineMoD 1000000007 - #defineMAX 0x7f7f7f7f to #definePI 3.14159265358979323 + #defineN 100004 - using namespacestd; thetypedefLong LongLL; * intCas,cass; $ intN,m,lll,ans;Panax Notoginseng LL Aans; - LL E[n],sum[n],l[n],r[n]; the LL A; + CharS[n]; A intMain () the { + #ifndef Online_judge - //freopen ("1.txt", "R", stdin); $ //freopen ("2.txt", "w", stdout); $     #endif -     inti,j,k; - LL x, y; the //for (scanf ("%d", &cass); cass;cass--) - //for (scanf ("%d", &cas), cass=1;cass<=cas;cass++)Wuyi //while (~scanf ("%s", s)) the      while(~SCANF ("%d",&N)) -     { Wuaans=0; -scanf"%s", s); Aboutn=strlen (s); $e[n]=1; e[n+1]=0; -          for(i=n-1; i;i--) e[i]= (e[i+1]*Ten)%MoD; -r[n]=1; r[n+1]=0; -          for(i=n-1; i+i>=n;i--) r[i]= (r[i+1]+e[i]* (n-i+1))%MoD; Al[1]=e[1];l[0]=0; +          for(i=2; i+i<=n+2; i++) l[i]= (l[i-1]+e[i]*i)%MoD; thesum[1]=e[1]; -          for(i=2; i<=n;i++) sum[i]=sum[i-1]+E[i]; $          the          for(i=1; i<=n;i++) the         { thea=s[i-1]-'0'; the             if(a==0)Continue; -x=i;y=n-i+1; in             if(x>y) Swap (x, y); theAans= (Aans+a*l[min (x, y)-1])%MoD; theAans= (Aans+a*r[max (x, y) +1])%MoD; AboutAans= (aans+ (a*x* (sum[y]-sum[x-1])) (%MOD)%MoD; theAans= (aans+ (e[i]*a)%mod* (x* (x1)/2+y* (y1)/2)%mod)%MoD; the         } theprintf"%lld\n", Aans); +     } -     return 0; the }Bayi /* the // the  - // - */

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"Mathematics" CSU 1810 Reverse (2016 Hunan province 12th session of computer Program design Competition)