1353. Milliard Vasya's Function
Time Limit: 1.0 second
Memory Limit: 16 MB
Vasya is the beginning mathematician. He decided to make an important contribution to the science and to become famous all over the world. But how can he do that if the most interesting facts such as Pythagor’s theorem are already proved? Correct! He is to think out something his own, original. So he thought out the Theory of Vasya’s Functions. Vasya’s Functions (VF) are rather simple: the value of the Nth VF in the point S is an amount of integers from 1 to N that have the sum of digits S. You seem to be great programmers, so Vasya gave you a task to find the milliard VF value (i.e. the VF with N = 109) because Vasya himself won’t cope with the task. Can you solve the problem?
Input
Output
The milliard VF value in the point S.
Sample
题意：输入s，输出自然数1...N中各位数字之和等于s的数的个数。
解题思路：
DP: F[i][j]表示i位数中，数字之和为j的数的个数，F[i][j] = F[i - 1][k]; (k = j....j-9 && k > 0);
递推方程由：F[i][j] = F[i - 1][j]个数尾部补0 + F[i - 1][j - k]个数头部补k，(k = 1...9, && j - k > 0)，而来。
Code:
#include<cstdio>
int F[10][82];
int count[82];
void init(){
int i = 0;
int j = 0;
int k = 0;
count[0] = 1;
for(i = 0; i < 10; i++){
for(j = 0; j < 82; j++){
F[i][j] = 0;
}
}
for(i = 0; i <= 9; i++){
F[1][i] = 1;
}
for(i = 2; i < 10; i++){
for(j = 1; j < 82; j++){
for(k = j; k > 0 && k >= j - 9; k--){
F[i][j] += F[i - 1][k];
}
}
}
count[1] = 10;
for(j = 2; j < 82; j++){
count[j] = 0;
for(i = 1; i < 10; i++){
count[j] += F[i][j];
}
}
}
int main(int argc, char* argv[], char* env[])
{
int n = 0;
init();
while(scanf("%d", &n) != EOF){
printf("%d\n", count[n]);
}
return 0;
}
posted on 2011-07-21 14:11
Lshain 阅读(400)
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