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题意:给定一个浮点数,让你在时间 t 内,变成一个最大的数,操作只有把某个小数位进行四舍五入,每秒可进行一次。
析:贪心策略就是从小数点开始找第一个大于等于5的,然后进行四舍五入,完成后再看看是不是还可以,一循环下去,直到整数位,或者没时间了。
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 <tr1/unordered_map> #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; using namespace std :: tr1; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const LL LNF = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 2e5 + 5; const int mod = 1e9 + 7; const int N = 1e6 + 5; const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1}; const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1}; const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); } 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; } char s[maxn]; int a[maxn], b[maxn]; int main(){ while(scanf("%d %d", &n, &m) == 2){ scanf("%s", s); int cnta = 0, cntb = 0; a[0] = b[0] = 0; bool ok = false; for(int i = 0; i < n; ++i){ if(s[i] == ‘.‘){ ok = true; continue; } if(ok) b[++cntb] = s[i] - ‘0‘; else a[++cnta] = s[i] - ‘0‘; } int cnt = 0; int y = 1, z = cntb; while(m--){ ok = false; bool x = true; for(int i = y; i <= z; ++i){ if(b[i] >= 5){ x = false; cnt = 1; ok = true; b[i] = -1; for(int j = i-1; j >= 0; --j){ if(b[j] + cnt > 9){ b[j] = -1; z = j; } else { b[j] += cnt; y = j; cnt = 0; break; } } } if(ok) break; } if(x) break; } if(b[0]){ cnt = 1; for(int j = cnta; j >= 0; --j){ if(a[j] + cnt > 9) a[j] = 0, cnt = 1; else { a[j] += cnt; cnt = 0; break; } } if(a[0]) printf("1"); for(int i = 1; i <= cnta; ++i) printf("%d", a[i]); } else{ int t = 0; for(int i = 1; i <= cntb; ++i) if(b[i] == -1){ t = i; break; } if(!t) t = cntb; for(int i = t; i > 0; --i) if(b[i] == 0 || b[i] == -1) b[i] = -1; else break; for(int i = 1; i <= cnta; ++i) printf("%d", a[i]); if(b[1] != -1){ printf("."); for(int i = 1; i <= cntb; ++i) if(b[i] == -1) break; else printf("%d", b[i]); } } printf("\n"); } return 0; }
CodeForces 718A Efim and Strange Grade (贪心)
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原文地址:http://www.cnblogs.com/dwtfukgv/p/5904412.html