HOJ 6433 Problem H. Pow
阿新 • • 發佈：2019-01-09
Problem H. PowTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)Total Submission(s): 1521 Accepted Submission(s): 721 Problem Description There are n numbers 3^0, 3^1, . . . , 3^n-1. Each time you can choose a subset of them (may be empty), and then add them up. Input The first line of the input contains an integer T , denoting the number of test cases. In each test case, there is a single integers n in one line, as described above. Output For each test case, output one line contains a single integer, denoting the answer. Sample Input 4 9 7 8 233 Sample Output 512 128 256 13803492693581127574869511724554050904902217944340773110325048447598592 |
知識點：string
#include<iostream> #include<cstdio> #include<string> using namespace std; string str[1010]; void init() { str[0] = "1"; bool flag = false; for(int i=1; i<=1000; i++) { string s = ""; for(int j=str[i-1].size()-1; j>=0; j--) { int t = (str[i-1][j]-'0')*2; if(flag) { t++; flag = false; } if(t<10) { s = (char)(t+'0') + s; }else { s = (char)(t-10+'0') + s; flag = true; } } if(flag) s = "1" + s; str[i] = s; flag = false; } } int main() { int T_T; init(); cin >> T_T; while(T_T--) { int n; cin >> n; cout << str[n] <<endl; } return 0; }