第一题:和谐数 /** * Created by dcp on 2018/09/16. */ /** * Created by dcp on 2018/09/16. */ const readline = require('readline'); const rl = readline.createInterface({ input: process.stdin, output: process.stdout }); var inputArr = []; rl.on('line', function (input) { inputArr.push(input); var nLine = +inputArr[0];//转化成数字类型 if (inputArr.length == (nLine + 1)) { var arr = inputArr.slice(1); for(var i=0;i<arr.length;i++){ hexie(arr[i]) } inputArr = []; } }); function hexie(n) { var sum=0; var t=parseInt(n); var arr=n.split('') var re=arr.map(function (ietm) { return ietm-0 }) for(var i=0;i<re.length;i++){ sum+=re[i]; } sum=parseInt(sum) if(t%sum===0){ console.log('Yes') }else { console.log('No') } } //第二题：坚果巧克力 /** * Created by dcp on 2018/9/16. */ /** * Created by dcp on 2018/09/16. */ var readline=require("readline"); var r1=readline.createInterface({ input:process.stdin, output:process.stdout }); //k控制输入行数 var k=3 //数组存储输入 var inputs=[]; r1.on("line",function(data){ inputs.push(data); if(inputs.length==2) { //字符串形式输出 var num0 = inputs[0].split(' ').toString() //数组形式输出 var num1 = inputs[1].split(' ') //将['2','3']转换成[2,3] // var num2=inputs[2].split(' ') // var mapResult=num2.map(function (item) { // return item-0 // }) jiangu(num1) } }); function jiangu(arr) { var cz=1,p=[]; for(var i=0;i<arr.length;i++){ if(arr[i]=='0'){ cz+=1; }else { p.push(cz) cz=1; } } p[0]=1; var s=1; for(var i=0;i<p.length;i++){ s*=p[i]; } console.log(s) } //第三题：能量矿石，通过0.4 #include <iostream> #include <fstream> using namespace std; int main() { #ifdef DEBUG ifstream cin("../Tencent/data4.in"); if (!cin.is_open()) { cout << "open error" << endl; exit(0); } #endif string S, P; cin >> S >> P; int lenP = P.length(); int lenS = S.length(); int idx = lenP; for (int i = 1, j = lenP - 1; i <= j; ++i, --j) { if (P.substr(0, i) == P.substr(j, lenP - j)) { idx = min(idx, j); } } int maxLen = 0; int preIdx = 0; int res = 0; for (int i = 0; i < lenS;) { if (S.substr(i, lenP) == P) { maxLen = max(maxLen, i + lenP - preIdx); i += idx; } else { // i+=lenP; ++i; preIdx = i; res += maxLen * maxLen; maxLen = 0; } } res += maxLen * maxLen; cout << res << endl; return 0; }