Vasya is the beginning mathematician. He decided to make an important contribution to the science an
2017-08-22 10:05
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题意:1~1000000000之间,各位数字之和等于给定s的数的个数
每行给出一个数s(1 ≤ s ≤ 81),求出1~10^9内各位数之和与s相等的数的个数。
1、只有s=1时,10^9的系数才能为1,否则就大于10^9;
所以和为1的要单一列出来。
2、如果s!=1:
定义状态dp[i][j]为前i位各位数之和为j的情况数量:对于前i为的数字之和最大为:9*i,即每一位数字都是9.
i=1、只有一位的数字,因为s>=1,所以最低位只能是1-9 其中的一个数字,
i>1、假设第i位放数字k(则k只能是0~9并且k<=s),若要使第前i位数字之和为j,那么前i-1位只能放j-k,由此得出动态转移方程:d[i]][j]=d[i][j]+d[i-1][j-k]
#include<cstdio>
int dp[9][81];
int main(){
int n;
for(int i=1;i<=9;i++)
dp[1][i]=1;
for(int i=1;i<=9;i++)
for(int j=1;j<=9*i;j++)
for(int k=0;k<=9 && k<=j;k++)
dp[i][j]+=dp[i-1][j-k];
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);
}
}
总结:关于数字每位数字和的问题。每一位只能是0~9的其中一个数字。最高位不能为零,所以最高位涉及到的数字就要单一列出来,作为特殊的处理。定义状态dp[i][j]为前i位各位数之和为j的情况数量,假设第i位放数字k(则k只能是0~9并且k<=s),若要使第前i位数字之和为j,那么前i-1位只能放j-k,由此得出动态转移方程:d[i]][j]=d[i][j]+d[i-1][j-k]
(0<=k<=j&&k<=9)
每行给出一个数s(1 ≤ s ≤ 81),求出1~10^9内各位数之和与s相等的数的个数。
1、只有s=1时,10^9的系数才能为1,否则就大于10^9;
所以和为1的要单一列出来。
2、如果s!=1:
定义状态dp[i][j]为前i位各位数之和为j的情况数量:对于前i为的数字之和最大为:9*i,即每一位数字都是9.
i=1、只有一位的数字,因为s>=1,所以最低位只能是1-9 其中的一个数字,
i>1、假设第i位放数字k(则k只能是0~9并且k<=s),若要使第前i位数字之和为j,那么前i-1位只能放j-k,由此得出动态转移方程:d[i]][j]=d[i][j]+d[i-1][j-k]
#include<cstdio>
int dp[9][81];
int main(){
int n;
for(int i=1;i<=9;i++)
dp[1][i]=1;
for(int i=1;i<=9;i++)
for(int j=1;j<=9*i;j++)
for(int k=0;k<=9 && k<=j;k++)
dp[i][j]+=dp[i-1][j-k];
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);
}
}
总结:关于数字每位数字和的问题。每一位只能是0~9的其中一个数字。最高位不能为零,所以最高位涉及到的数字就要单一列出来,作为特殊的处理。定义状态dp[i][j]为前i位各位数之和为j的情况数量,假设第i位放数字k(则k只能是0~9并且k<=s),若要使第前i位数字之和为j,那么前i-1位只能放j-k,由此得出动态转移方程:d[i]][j]=d[i][j]+d[i-1][j-k]
(0<=k<=j&&k<=9)
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