URAL 1353 Milliard Vasya's Function(DP)
2014-07-24 10:25
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Milliard Vasya's Function
大意:
求1-10^9之间的数中,各数位和为s的数的个数。
思路:
dp[i][j]表示位数是i的数字各个位之和为j的数的个数(1<= i <=9, 1<= j <=81)。
先DP出1到9 位数上各位之和的个数,(dp[i][j] = dp[i-1][j]+sum(dp[i-1][j-1], dp[i-1][j-2]......., dp[i-1][j-9]),dp[i-1][j]表示位数少一位的时候在末位加0), 输入s的时候再求一个累加和。
#include <stdio.h> int dp[10][100]; int main() { for(int i = 1; i <= 9; ++i){ dp[1][i] = 1; } for(int i = 2; i <= 9; ++i){ for(int j = 1; j <= 81; ++j){ dp[i][j] = dp[i-1][j]; for(int k = 1; k <= 9 && j-k > 0; ++k){ dp[i][j] += dp[i-1][j-k]; } } } int n; while(~scanf("%d", &n)){ if(n == 1){ printf("10\n"); continue; } int sum = 0; for(int i = 1; i <= 9; ++i){ sum += dp[i] ; } printf("%d\n", sum); } return 0; }
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