题意:给定 n 首诗,然后让你给他打分,有四种规则,第一种是按音节打,如果音节在8-12之间,就加10,第二种就是按押韵打,如果13,24一个押韵就加20分。
第三种按音节数打,13,24有一个一样的就加10分,第四种是扣分的,如果多于4段,扣10分。
析:按照他说的来,就好,这个输入就挺恶心的,好难处理,而且一种很变态的数据就是标点后面没有空格,你还得自己去判断。。。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #define frer freopen("in.txt", "r", stdin) #define frew freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e2 + 5; const int mod = 1e9 + 7; const char *mark = "+-*"; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline int Min(int a, int b){ return a < b ? a : b; } inline int Max(int a, int b){ return a > b ? a : b; } inline LL Min(LL a, LL b){ return a < b ? a : b; } inline LL Max(LL a, LL b){ return a > b ? a : b; } inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int a[500]; bool b[500]; bool c[500]; vector<string> v[500]; int ans[10]; int main(){ int T; cin >> T; for(int kase = 1; kase <= T; ++kase){ int cnt = 0; string s; for(int i = 0; i < 500; ++i) v[i].clear(); while(true){ cin >> s; bool ok = false, ok1 = false; if(s[s.size()-1] == ',') s.resize(s.size()-1), ok1 = true; if(s[s.size()-1] == '.') ok = true, s.resize(s.size()-1); for(int i = 0; i < s.size(); ++i) if(s[i] == ',') s[i] = ' '; stringstream ss(s); int x = 0; while(ss >> s){ if(x) ++cnt; v[cnt].push_back(s); ++x; } if(ok1) ++cnt; if(ok) break; } memset(ans, 0, sizeof ans); memset(a, 0, sizeof a); for(int i = 0; i <= cnt; ++i){ a[i] = 0; for(int j = 0; j < v[i].size(); ++j){ int len = v[i][j].size(); if(len == 6){ s = v[i][j]; bool ok = false; for(int k = 0; k < s.size(); ++k){ if(k+1 < s.size() && s[k] == 'n' && s[k+1] == 'g'){ a[i] += 2; ok = true; break; } else if(k+1 < s.size() && s[k] == 'n' && s[k+1] == 'y'){ a[i] += 2; ok = true; break; } } if(!ok) a[i] += 3; } else if(len > 6) a[i] += 3; else if(len == 4 || len == 5) a[i] += 2; else if(len == 3){ s = v[i][j]; if(towlower(s[0]) == 'a' || towlower(s[0]) == 'e' || towlower(s[0]) == 'o' || towlower(s[0]) == 'i' || towlower(s[0]) == 'u') a[i] += 2; else ++a[i]; } else ++a[i]; } } for(int i = 0; i < 4; ++i) if(a[i] >= 8 && a[i] <= 12) ans[1] += 10; if(cnt >= 3){ string s1 = v[0][v[0].size()-1], s2 = v[1][v[1].size()-1], s3 = v[2][v[2].size()-1], s4 = v[3][v[3].size()-1]; if(s1[s1.size()-2] == s3[s3.size()-2] && s1[s1.size()-1] == s3[s3.size()-1]) ans[2] += 20; if(s2[s2.size()-2] == s4[s4.size()-2] && s2[s2.size()-1] == s4[s4.size()-1]) ans[2] += 20; } else if(cnt == 2){ string s1 = v[0][v[0].size()-1], s2 = v[1][v[1].size()-1], s3 = v[2][v[2].size()-1]; if(s1[s1.size()-2] == s3[s3.size()-2] && s1[s1.size()-1] == s3[s3.size()-1]) ans[2] += 20; } if(a[2] && a[0] == a[2]) ans[3] += 10; if(a[1] && a[3] && a[1] == a[3]) ans[3] += 10; for(int i = 4; i <= cnt; ++i) ans[4] += 10; ans[5] = ans[1] + ans[2] + ans[3] - ans[4]; printf("Case #%d: %d %d %d %d %d\n", kase, ans[1], ans[2], ans[3], ans[4], ans[5]); } return 0; }