ICPC 6823 Counting substhreengs
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题意:求出满足要求的子串个数,要求是:子串都是数字,并且能被3整除。
做法:动态规划的思想。弄一个数组记录当前选定值之前的情况,num[0]记录余数为0的个数。假设选定当前值作为数字的最后一个位数,如果当前值mod3等于1,那么前面num[2]的数加上当前值都能变成余数0,所以num[0]=num[2].以此类推......
不用long long 会wa。
#include<iostream>#include<string>#include<cstring>#include<algorithm>using namespace std;int main(){string s;while(cin >> s){long long num[4] = {0};long long ans = 0;for(long long i=0;i<s.length();i++){if(s[i]>='0'&&s[i]<='9'){long long x = (s[i]-'0')%3;long long x1,x2,x3;if(x==0){x1 = num[0]+1;x2 = num[1];x3 = num[2];}else if(x==1){x1 = num[2];x2 = num[0]+1;x3 = num[1];}else if(x==2){x1 = num[1];x2 = num[2];x3 = num[0]+1;}num[0] = x1;num[1] = x2;num[2] = x3;ans+=num[0];}else{num[0] = 0;num[1] = 0;num[2] = 0;}}cout<<ans<<endl;}return 0;}
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