Submission #1800752

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Source Code Expand

/*
神题
1.b[i]=a[i]~a[n*2-i]
2.b中i之前所有数都不能在b[i]~b[i+1]之间（不包括端点）
f[i][j][k]=填完了b的后i个数，b的倒数第i个数有j种选择，它在这j种选择中排第k小，方案数
转移时计算选择种类数，枚举当前选择哪种
O(n^4)
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 110
#define mod 1000000007
using namespace std;
int n, w[N], f[N][N][N], sum, ans, now, x, y, x1, x2;
int main(){
	scanf("%d", &n); for(int i=1; i<=n*2-1; i++)scanf("%d", &w[i]);
	sort(w+1, w+n*2);
	sum=1; for(int i=2; i<=n*2-1; i++)if(w[i-1]<w[i])sum++;
	memset(f, 0, sizeof(f)); f[1][1][1]=1; now=1;
	for(int i=1; i<=n-1; i++){
		x1=w[n-i+1]==w[n-i]?0:1;
		x2=w[n+i-1]==w[n+i]?0:1;
		for(int j=1; j<=now; j++)
			for(int k=1; k<=j; k++){
				x=j+x1+x2; y=k+x1;
				for(int ii=1; ii<=y-1; ii++)f[i+1][x-y+ii+1][ii]=(f[i+1][x-y+ii+1][ii]+f[i][j][k])%mod;
				f[i+1][x][y]=(f[i+1][x][y]+f[i][j][k])%mod;
				for(int ii=y+1; ii<=x; ii++)f[i+1][x-ii+y+1][y+1]=(f[i+1][x-ii+y+1][y+1]+f[i][j][k])%mod;
			}
		now+=x1+x2;
	}
	ans=0;
	for(int i=1; i<=sum; i++)
		for(int j=1; j<=i; j++)ans=(ans+f[n][i][j])%mod;
	printf("%d", ans);
}

Submission Info

Compile Error

./Main.cpp: In function ‘int main()’:
./Main.cpp:17:17: warning: ignoring return value of ‘int scanf(const char*, ...)’, declared with attribute warn_unused_result [-Wunused-result]
  scanf("%d", &n); for(int i=1; i<=n*2-1; i++)scanf("%d", &w[i]);
                 ^
./Main.cpp:17:64: warning: ignoring return value of ‘int scanf(const char*, ...)’, declared with attribute warn_unused_result [-Wunused-result]
  scanf("%d", &n); for(int i=1; i<=n*2-1; i++)scanf("%d", &w[i]);
                                                                ^

Judge Result