HUST 1599 - Multiple(动态规划)

1599 - Multiple

时间限制：2秒 内存限制：64兆
461 次提交 111 次通过
题目描述
Rocket323 loves math very much. One day, Rocket323 got a number string. He could choose some consecutive digits from the string to form a number. Rocket323 loves 64 very much, so he wanted to know how many ways can he choose from the string so the number he got multiples of 64 ?

A number cannot have leading zeros. For example, 0, 64, 6464, 128 are numbers which multiple of 64 , but 12, 064, 00, 1234 are not.
输入
Multiple cases, in each test cases there is only one line consist a number string.
Length of the string is less than 3 * 10^5 .

Huge Input , scanf is recommended.
输出
Print the number of ways Rocket323 can choose some consecutive digits to form a number which multiples of 64.
样例输入
64064
样例输出
5
提示
There are five substrings which multiples of 64.
[64]064
640[64]
64[0]64
[64064]
[640]64
来源
Problem Setter : Yang Xiao

思路：
根据同余定理，每次枚举到一个数，都把前面的存在的余数乘以10加上这个数再对64取余，就产生新的余数，最后统计余数是0的个数有多少

#include <iostream>#include <string.h>#include <math.h>#include <algorithm>#include <stdio.h>#include <stdlib.h>using namespace std;char a[300005];long long int dp[2][65];long long int tag[65];long long int ans;long long int res;int now;int main(){    while(scanf("%s",a)!=EOF)    {        ans=0;res=0;        int len=strlen(a);        memset(dp,0,sizeof(dp));        now=0;        for(int i=0;i<len;i++)        {            for(int j=63;j>=0;j--)                dp[now^1][j]=0;            for(int j=63;j>=0;j--)            {                int num=(j*10+a[i]-'0'+64)%64;                dp[now^1][num]+=dp[now][j];            }            if(a[i]!='0')                dp[now^1][a[i]-'0']++;            else                ans++;            res+=dp[now^1][0];            now^=1;        }        printf("%lld\n",res+ans);    }    return 0;}