3107: [cqoi2013]二进制a+b
Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 995 Solved: 444
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Description
输入三个整数a, b, c,把它们写成无前导0的二进制整数。比如a=7, b=6, c=9,写成二进制为a=111, b=110, c=1001。接下来以位数最多的为基准,其他整数在前面添加前导0,使得a, b, c拥有相同的位数。比如在刚才的例子中,添加完前导0后为a=0111, b=0110, c=1001。最后,把a, b, c的各位进行重排,得到a’, b’, c’,使得a’+b’=c’。比如在刚才的例子中,可以这样重排:a’=0111, b’=0011, c’=1010。
你的任务是让c’最小。如果无解,输出-1。
Input
输入仅一行,包含三个整数a, b, c。
Output
输出仅一行,为c’的最小值。
Sample Input
7 6 9
Sample Output
10
HINT
a,b,c<=2^30
题解:
参考代码:dp[i][j][k][l][m]表示:第I位,第1,2,3个数分别用了几个一,m表示是否有进位:
分类讨论:
对于m=0的情况:
dp[i+1][j+1][k+1][l+1][1]=min(dp[i+1][j+1][k+1][l+1][1],tmp+(1<<i+1)); dp[i+1][j+1][k][l+1][0]=min(dp[i+1][j+1][k][l+1][0],tmp+(1<<i)); dp[i+1][j][k+1][l+1][0]=min(dp[i+1][j][k+1][l+1][0],tmp+(1<<i)); dp[i+1][j][k][l][0]=min(dp[i+1][j][k][l][0],tmp);
对于m=1的情况:
dp[i+1][j+1][k+1][l+1][1]=min(dp[i+1][j+1][k+1][l+1][1],tmp+(1<<i+1)); dp[i+1][j][k+1][l][1]=min(dp[i+1][j][k+1][l][1],tmp+(1<<i)); dp[i+1][j+1][k][l][1]=min(dp[i+1][j+1][k][l][1],tmp+(1<<i)); dp[i+1][j][k][l][0]=min(dp[i+1][j][k][l][0],tmp);
#include<bits/stdc++.h> using namespace std; #define clr(a,b) memset(a,b,sizeof(a)) typedef long long ll; const ll inf=0x3f3f3f3f3f3f3f3fll; ll a,b,c; ll la,lb,lc,len; ll dp[62][40][40][40][2]; void getlen(){len=max(log2(a)+1,max(log2(b)+1,log2(c)+1));} ll getnum(ll x) { ll sum=0; for(int i=0;i<62;++i) {if( x & (1ll<<i) ) sum++; } return sum; } void work() { dp[0][0][0][0][0]=0; for(ll i=0;i<len;++i) { for(ll j=0;j<=la;++j) { for(ll k=0;k<=lb;++k) { for(ll l=0;l<=lc;++l) { long long tmp=dp[i][j][k][l][0];//枚举最后一位不进位的情况 dp[i+1][j+1][k+1][l+1][1]=min(dp[i+1][j+1][k+1][l+1][1],tmp+(1<<i+1)); dp[i+1][j+1][k][l+1][0]=min(dp[i+1][j+1][k][l+1][0],tmp+(1<<i)); dp[i+1][j][k+1][l+1][0]=min(dp[i+1][j][k+1][l+1][0],tmp+(1<<i)); dp[i+1][j][k][l][0]=min(dp[i+1][j][k][l][0],tmp); tmp=dp[i][j][k][l][1];//枚举最后一位进位的情况 dp[i+1][j+1][k+1][l+1][1]=min(dp[i+1][j+1][k+1][l+1][1],tmp+(1<<i+1)); dp[i+1][j][k+1][l][1]=min(dp[i+1][j][k+1][l][1],tmp+(1<<i)); dp[i+1][j+1][k][l][1]=min(dp[i+1][j+1][k][l][1],tmp+(1<<i)); dp[i+1][j][k][l][0]=min(dp[i+1][j][k][l][0],tmp); } } } } //cout<<len<<" "<<la<<" "<<lb<<" "<<lc<<endl; if(dp[len][la][lb][lc][0]>=inf) printf("-1\n"); else printf("%d\n",dp[len][la][lb][lc][0]); } int main() { scanf("%lld%lld%lld",&a,&b,&c); getlen(); memset(dp,inf,sizeof dp); la=getnum(a);lb=getnum(b);lc=getnum(c); work(); return 0; }