0%

ZOJ - 3950 How Many Nines (模拟

题目链接:
https://cn.vjudge.net/problem/ZOJ-3950

题目大意:
题目给出两个日期a与b,统计从日期a到b之间的所有日期里共有多有个’9’

分析:
分成四部分写,年份不同,月份不同,日期不同和完全相同,前缀和处理各种数据,最后重点处理下年份不同的情况即可

代码:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <time.h>
using namespace std;


int y1,m1,d1,y2,m2,d2;

int year = 65;
int day[2][20] = {{0,31,28,31,30,31,30,31,31,30,31,30,31},{0,31,29,31,30,31,30,31,31,30,31,30,31}};
int mart[2][30][50];
int lp[12000];
int np[12000];
int mnine[2][30][50];
int nine[12000];

inline int count9(int n)
{
int cnt = 0;
while (n)
{
if (n%10==9)
cnt++;
n/=10;
}
return cnt;
}

bool leap(int n)
{
return (!(n%4)&&(n%100))||!(n%400);
}

void init()
{
for (int i = 1 ; i <= 10000 ; i ++)
nine[i] = count9(i);
int cnt = 0;
int sum = 0;
for (int i = 1 ; i <= 12 ; i ++)
{
for (int j = 1 ; j <= day[0][i] ; j ++)
{
mart[0][i][j] = cnt++;
mnine[0][i][j] = sum;
sum += nine[i]+nine[j];
}
}
cnt = 0,sum = 0;
for (int i = 1 ; i <= 12 ; i ++)
{
for (int j = 1 ; j <= day[1][i] ; j ++)
{
mart[1][i][j] = cnt++;
mnine[1][i][j] = sum;
sum += nine[i]+nine[j];
}
}
cnt = 0;
lp[2000] = 0;
np[2000] = 0;
for (int i = 2001 ; i <= 9999 ; i ++)
{
bool flag = leap(i-1);
lp[i] = flag+lp[i-1];
np[i] = np[i-1] + nine[i-1]*(365+flag);
}

return;
}

int main()
{
init();
int T;
//cout<<mart[1][12][31]<<endl;
scanf("%d",&T);
clock_t st,ed;
while (T--){
int ans = 0;
scanf("%d%d%d",&y1,&m1,&d1);
scanf("%d%d%d",&y2,&m2,&d2);
// for (int i = y1+1 ; i < y2 ; i++)
// ans += year + leap(i) + nine[i]*(365+leap(i));
bool flag = leap(y1);
int cnt = 0 ;
if (y1==y2)
{
if (m1==m2)
{
if (d1==d2)
ans += nine[y1]+nine[m1]+nine[d1];
else
{
ans += mnine[flag][m2][d2] - mnine[flag][m1][d1]+nine[m2]+nine[d2];
cnt = mart[flag][m2][d2] - mart[flag][m1][d1]+1;
ans += nine[y1]*cnt;
}
}
else
{
ans += mnine[flag][m2][d2] - mnine[flag][m1][d1]+nine[m2]+nine[d2];
cnt = mart[flag][m2][d2] - mart[flag][m1][d1]+1;
ans += nine[y1]*cnt;
}
}
else
{
ans += np[y2] -np[y1+1];
ans += (y2-y1-1)*65 + lp[y2]-lp[y1+1];
ans += mnine[flag][12][31]-mnine[flag][m1][d1];
cnt = 365 + flag - mart[flag][m1][d1];
ans += nine[y1]*cnt;
cnt = 0;
flag = leap(y2);
ans += mnine[flag][m2][d2]+nine[m2]+nine[d2];
cnt = mart[flag][m2][d2]+1;
ans += nine[y2]*cnt;
}
printf("%d\n",ans);
}
return 0;
}