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uva1374快速冪計算

2019-11-08 01:50:56
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來源:轉載
供稿:網(wǎng)友

挺久沒寫博客了,這道題可以打一張1到1000的表,生成表大約需要8秒,但是生成后可以把結果復制粘貼到一個數(shù)組中,使效率達到最大化,當然不打表一個一個求也不會超時,代碼如下

#include <bits/stdc++.h>using namespace std;//int n, maxd, current[1005];////bool dfs(int cur, int d)//{//    if (d >= maxd) {//        if (cur == n) return true;//        else return false;//    }//    for (int i = d; i >= 0; i--) {//        int MAX = 0;//        for (int j = 0; j <= d; j++) {//            if (current[j] > MAX) MAX = current[j];//        }//        if ((cur + MAX) << (maxd - d - 1) < n) return false;////        current[d + 1] = cur + current[i];//        if (dfs(cur + current[i], d + 1)) return true;////        current[d + 1] = cur - current[i];//        if (dfs(cur - current[i], d + 1)) return true;//    }//    return false;//}////int main()//{//    freopen("out.txt", "w", stdout);//    int save[1001] = {0};//    for (n = 2; n <= 1000; n++) {//        for (maxd = 1; ; maxd++) {//            memset(current, 0, sizeof(current));//            current[0] = 1;//            if (dfs(1, 0)) break;//        }//        save[n] = maxd;//    }////    for (int i = 0; i < 1001; i++) {//        cout << save[i] << ',';//        if (i % 20 == 0) cout << endl;//    }//    return 0;//}int save[] = {0,0,1,2,2,3,3,4,3,4,4,5,4,5,5,5,4,5,5,6,5,6,6,6,5,6,6,6,6,7,6,6,5,6,6,7,6,7,7,7,6,7,7,7,7,7,7,7,6,7,7,7,7,8,7,8,7,8,8,8,7,8,7,7,6,7,7,8,7,8,8,8,7,8,8,8,8,8,8,8,7,8,8,8,8,8,8,9,8,9,8,9,8,8,8,8,7,8,8,8,8,9,8,9,8,9,9,9,8,9,9,9,8,9,9,9,9,9,9,9,8,9,9,9,8,9,8,8,7,8,8,9,8,9,9,9,8,9,9,9,9,9,9,9,8,9,9,9,9,9,9,10,9,9,9,9,9,9,9,9,8,9,9,9,9,9,9,10,9,10,9,10,9,10,10,10,9,10,10,10,9,10,10,10,9,10,9,10,9,9,9,9,8,9,9,9,9,10,9,10,9,10,10,10,9,10,10,10,9,10,10,10,10,10,10,10,9,10,10,10,10,10,10,10,9,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,9,10,10,10,10,10,10,10,9,10,10,10,9,10,9,9,8,9,9,10,9,10,10,10,9,10,10,11,10,11,10,10,9,10,10,11,10,11,10,10,10,10,10,10,10,10,10,10,9,10,10,10,10,10,10,11,10,10,10,11,10,11,11,11,10,11,10,11,10,11,10,11,10,11,10,10,10,10,10,10,9,10,10,10,10,10,10,11,10,11,10,11,10,11,11,11,10,11,11,11,10,11,11,11,10,11,11,11,11,11,11,11,10,11,11,11,11,11,11,11,10,11,11,11,11,11,11,11,10,11,11,11,10,11,11,11,10,11,10,11,10,10,10,10,9,10,10,10,10,11,10,11,10,11,11,11,10,11,11,11,10,11,11,11,11,11,11,11,10,11,11,11,11,11,11,11,10,11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,10,11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,10,11,11,11,11,11,11,11,11,11,11,11,11,12,11,12,11,11,11,12,11,12,11,11,11,11,11,11,11,11,11,11,10,11,11,11,11,11,11,11,11,11,11,12,11,12,11,11,10,11,11,12,11,12,11,11,10,11,11,11,10,11,10,10,9,10,10,11,10,11,11,11,10,11,11,12,11,12,11,11,10,11,11,12,11,12,12,11,11,12,12,12,11,12,11,11,10,11,11,12,11,12,12,12,11,11,12,12,11,12,11,12,11,11,11,12,11,12,11,11,11,12,11,11,11,11,11,11,10,11,11,11,11,11,11,12,11,11,11,12,11,12,12,12,11,12,11,12,11,12,12,12,11,12,12,12,12,12,12,12,11,12,12,12,11,12,12,12,11,12,12,12,11,12,12,12,11,12,12,12,11,12,11,12,11,12,11,11,11,11,11,11,10,11,11,11,11,11,11,12,11,12,11,12,11,12,12,12,11,12,12,12,11,12,12,12,11,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,11,12,12,12,11,12,12,12,11,12,11,12,11,11,11,11,10,11,11,11,11,12,11,12,11,12,12,12,11,12,12,12,11,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,12,12,12,13,12,12,12,12,11,12,12,12,12,13,12,12,12,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,12,12,12,12,12,12,12,13,12,12,12,13,12,13,12,12,12,13,12,12,12,12,12,12,11,12,12,12,12,12,12,13,12,12,12,13,12,13,12,12,12,13,12,13,12,13,12,12,12,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,12,12,12,13,12,13,12,13,12,13,12,13,12,13,12,13,12,13,13,13,12,13,13,13,12,13,12,13,12,13,13,13,12,13,13,13,12,13,12,13,12,12,12,13,12,13,12,12,12,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,12,12,12,13,12,13,12,12,12,12,12,13,12,13,13,13,12,13,13,13,12,13,12,12,11,12,12,13,12,13,13,13,12};int main(){    int n;    while (cin >> n && n) {        cout << save[n] << endl;    }}


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