| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 108487 | Accepted: 33679 |
Description
Input
当p = e = i = d = -1时,输入数据结束。
Output
采用以下格式:
Case 1: the next triple peak occurs in 1234 days.
注意:即使结果是1天,也使用复数形式“days”。
Sample Input
0 0 0 0 0 0 0 100 5 20 34 325 4 5 6 7 283 102 23 320 203 301 203 40 -1 -1 -1 -1
Sample Output
Case 1: the next triple peak occurs in 21252 days. Case 2: the next triple peak occurs in 21152 days. Case 3: the next triple peak occurs in 19575 days. Case 4: the next triple peak occurs in 16994 days. Case 5: the next triple peak occurs in 8910 days. Case 6: the next triple peak occurs in 10789 days.
思路:
中国剩余定理,解释见代码中
参考:http://blog.youkuaiyun.com/y11201/article/details/8623930
代码:
#include <stdio.h>
int R1, R2, R3, R;
int A = 23, B = 28, C = 33;
void f(){
R = A * B * C;
R1 = B * C;
for(int i = 1; ; i ++){ // 保证R1能整除B、C,并且R1/A == 1。(之后kR1仍能整除 B、C , kR1/A == k),下同
if(R1 * i % A == 1){ // 此步目的在于保证 k*R1 仍能整除B\C, k*R1/A = k *1
R1 *= i;
break;
}
}
R2 = A * C;
for(int i = 1; ; i ++){
if(R2 * i % B == 1){
R2 *= i;
break;
}
}
R3 = A * B;
for(int i = 1; ; i ++){
if(R3 * i % C == 1){
R3 *= i;
break;
}
}
// printf("R1 = %d\n", R1);
// printf("R2 = %d\n", R2);
// printf("R3 = %d\n", R3);
// printf("R = %d\n", R);
}
int main()
{
f();
int a, b, c, day;
for(int i = 1; ; i ++){
scanf("%d%d%d%d", &a, &b, &c, &day);
if(a == -1){
break;
}
int res = (R1 * a + R2 * b + R3 * c - day) % R; // R2、R3 能整除 A,则R2+R3也能整除A ; 另外R1/A==1, 则(R1+R2+R2)/A = 1,(aR1+bR2+cR3)/A=a;
res = (res + R - 1) % R + 1; // 如果res是负数,变为正数, 先-1的原因是保证res>0 (res +R >= 0), 是0时代表增加一个周期
printf("Case %d: the next triple peak occurs in %d days.\n", i, res);
}
return 0;
}
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