CodeForces 718A Efim and Strange Grade (贪心)

Posted 2020-08-10 dwtfukgv

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;
}