bop_2014_round_2b

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1. 神奇的数列

仔细观察题目对数列的要求，首先确定1和2表示不出来，因为能表示的最小数是3，之后由于题目要求如果有多个数列，输出任意一个即可，所以直接构造即可。

#include <stdio.h>typedef unsigned long long ULL;int main() {  int T;  ULL N;  scanf("%d", &T);  for (int c = 1; T--; ++c) {    scanf("%llu", &N);    if (N < 3) {      printf("Case %d: no solution\n", c);      continue;    }    ULL a1, a2, a3;    a3 = 1; a2 = N - a3; a1 = a2 * 2;    printf("Case %d: %llu %llu %llu\n", c, a1, a2, a3);  }  return 0;}

2. 字符串压缩

先将串恢复到游程编码之前，之后计算行数，根据行数将串拆分成行存储在vector中，对vector排序比较即可

#include <iostream>#include <sstream>#include <string>#include <vector>#include <algorithm>using namespace std;typedef unsigned ULL;void Split2Rows(string &raw, vector<string> &rows, int L) {  rows.clear();  int cl = raw.length() / L;  string row;  for (int i = 0; i < cl; ++i) {    row.clear();    for (int j = i, k = 0; k < L; j += cl, ++k) {      row += raw[j];    }    rows.push_back(row);  }  return;}void Recover(string &str, string &raw) {  raw.clear();  stringstream ss(str);  int num;  char tmp;  ss >> num >> tmp; //stream只有读超过末尾才会处于错误状态，最后读到的是脏数据  while (ss) {    for (int i = 0; i < num; raw += tmp, ++i) {}    ss >> num >> tmp;  }  return;}int main() {  int T, L;  string str1, str2, raw1, raw2;  vector<string> rows1, rows2;  cin >> T;  for (int c = 1; T--; ++c) {    cin >> L >> str1 >> str2;    Recover(str1, raw1);    Recover(str2, raw2);    if (raw1.length() != raw2.length()) {      cout << "Case " << c << ": No" << endl;      continue;    }    Split2Rows(raw1, rows1, L);    Split2Rows(raw2, rows2, L);    sort(rows1.begin(), rows1.end());    sort(rows2.begin(), rows2.end());    if (rows1 == rows2)      cout << "Case " << c << ": Yes" << endl;    else      cout << "Case " << c << ": No" << endl;  }  return 0;}

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