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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