【SSE

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【SSE

##Question
满足特异条件的数列。输入m和n（20≥m≥n≥0），求出满足以下方程式的正整数数列i1,i2,…,in，使得i1+i2+…+in=m，且i1≥i2≥…≥in。例如：
当n=4，m=8时，将得到如下5个数列：
5 1 1 1 4 2 1 1 3 3 1 1 3 2 2 1 2 2 2 2
**输入格式要求：“%d” 提示信息：“Please enter requried terms (<=10):”
" their sum:"
**输出格式要求：“There are following possible series:\n” “[%d]:” “%d”
程序运行示例1：
Please enter requried terms (<=10): 4 8
their sum:There are following possible series:
[1]:5111
[2]:4211
[3]:3311
[4]:3221
[5]:2222
程序运行示例2：
Please enter requried terms (<=10):4 10
their sum:There are following possible series:

##Code

#include <stdio.h>int num[11];int count = 0;
void dfs(int n, int now, int m, int max){if(now==n){if(max>=m&&m>=1){num[n] = m;count++;printf("[%d]:",count);for(int i=1; i<=n; i++){printf("%d",num[i]);}printf("\n");}return;}for(int i=max; i>=m/(n-now+1); i--){num[now] = i;dfs(n, now+1, m-i, i);}
}int main(){printf("Please enter requried terms (<=10):");int n,m;scanf("%d %d",&n, &m);printf("                             their sum:");printf("There are following possible series:\n");dfs(n,1,m,m-n+1);
}

【SSE

##Question
满足特异条件的数列。输入m和n（20≥m≥n≥0），求出满足以下方程式的正整数数列i1,i2,…,in，使得i1+i2+…+in=m，且i1≥i2≥…≥in。例如：
当n=4，m=8时，将得到如下5个数列：
5 1 1 1 4 2 1 1 3 3 1 1 3 2 2 1 2 2 2 2
**输入格式要求：“%d” 提示信息：“Please enter requried terms (<=10):”
" their sum:"
**输出格式要求：“There are following possible series:\n” “[%d]:” “%d”
程序运行示例1：
Please enter requried terms (<=10): 4 8
their sum:There are following possible series:
[1]:5111
[2]:4211
[3]:3311
[4]:3221
[5]:2222
程序运行示例2：
Please enter requried terms (<=10):4 10
their sum:There are following possible series:

##Code

#include <stdio.h>int num[11];int count = 0;
void dfs(int n, int now, int m, int max){if(now==n){if(max>=m&&m>=1){num[n] = m;count++;printf("[%d]:",count);for(int i=1; i<=n; i++){printf("%d",num[i]);}printf("\n");}return;}for(int i=max; i>=m/(n-now+1); i--){num[now] = i;dfs(n, now+1, m-i, i);}
}int main(){printf("Please enter requried terms (<=10):");int n,m;scanf("%d %d",&n, &m);printf("                             their sum:");printf("There are following possible series:\n");dfs(n,1,m,m-n+1);
}

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