作业帮要求和程序如下,不知道错在哪里,Uncle Da is very smart.His IQ was just 0 on his birthday.But it grows 1 everyday.For example,when Uncle Da is 1 year old,his IQ is 365!Ziyao big god is smarter then Uncle Da.His IQ was 1 on his birthday,and
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/08 03:02:15
要求和程序如下,不知道错在哪里,Uncle Da is very smart.His IQ was just 0 on his birthday.But it grows 1 everyday.For example,when Uncle Da is 1 year old,his IQ is 365!Ziyao big god is smarter then Uncle Da.His IQ was 1 on his birthday,and
要求和程序如下,不知道错在哪里,
Uncle Da is very smart.His IQ was just 0 on his birthday.But it grows 1 everyday.For example,
when Uncle Da is 1 year old,his IQ is 365!
Ziyao big god is smarter then Uncle Da.His IQ was 1 on his birthday,and grows 1 time every day.
For example,when Ziyao is 1 year old,his IQ is 2^365!
Uncle Da's IQ is x when he is n years old.Ziyao's IQ is y when he is m years old.
(We assume that every year is 365 days)
Now a martian want to know how much is (y-x)%400009.
【input format】
Two numbers,n and m.(x,y
要求和程序如下,不知道错在哪里,Uncle Da is very smart.His IQ was just 0 on his birthday.But it grows 1 everyday.For example,when Uncle Da is 1 year old,his IQ is 365!Ziyao big god is smarter then Uncle Da.His IQ was 1 on his birthday,and
虽然不知道哪里错了,但我这有个运行正确的.
int n, m, i, j, sum = 0;
scanf("%d%d", &n, &m);
int x = (365 * n) % 400009 , y = 1;
for (j = 1; j 400009) y -= 400009;
}
printf("%d\n", (y - x + 400009) % 400009);