POJ 2506 Tiling(高精度+递推)

高精度模版(bin神的)

/* * 高精度,,支持乘法和加法 */struct BigInt{const static int mod = 10000;const static int DLEN = 4;int a[600],len;BigInt(){memset(a,0,sizeof(a));len = 1;}BigInt(int v){memset(a,0,sizeof(a));len = 0;do{a[len++] = v%mod;v /= mod;}while(v);}BigInt(const char s[]){memset(a,0,sizeof(a));int L = strlen(s);len = L/DLEN;if(L%DLEN)len++;int index = 0;for(int i = L-1;i >= 0;i -= DLEN){int t = 0;int k = i – DLEN + 1;if(k < 0)k = 0;for(int j = k;j <= i;j++)t = t*10 + s[j] – '0';a[index++] = t;}}BigInt operator +(const BigInt &b)const{BigInt res;res.len = max(len,b.len);for(int i = 0;i <= res.len;i++)res.a[i] = 0;for(int i = 0;i < res.len;i++){res.a[i] += ((i < len)?a[i]:0)+((i < b.len)?b.a[i]:0);res.a[i+1] += res.a[i]/mod;res.a[i] %= mod;}if(res.a[res.len] > 0)res.len++;return res;}BigInt operator *(const BigInt &b)const{BigInt res;for(int i = 0; i < len;i++){int up = 0;for(int j = 0;j < b.len;j++){int temp = a[i]*b.a[j] + res.a[i+j] + up;res.a[i+j] = temp%mod;up = temp/mod;}if(up != 0)res.a[i + b.len] = up;}res.len = len + b.len;while(res.a[res.len – 1] == 0 &&res.len > 1)res.len–;return res;}void output(){printf("%d",a[len-1]);for(int i = len-2;i >=0 ;i–)printf("%04d",a[i]);printf("\n");}};AC代码:

//844K0MS#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>using namespace std;/* * 高精度,支持乘法和加法 */struct BigInt{const static int mod = 10000;const static int DLEN = 4;int a[600],len;BigInt(){memset(a,0,sizeof(a));len = 1;}BigInt(int v){memset(a,0,sizeof(a));len = 0;do{a[len++] = v%mod;v /= mod;}while(v);}BigInt(const char s[]){memset(a,0,sizeof(a));int L = strlen(s);len = L/DLEN;if(L%DLEN)len++;int index = 0;for(int i = L-1;i >= 0;i -= DLEN){int t = 0;int k = i – DLEN + 1;if(k < 0)k = 0;for(int j = k;j <= i;j++)t = t*10 + s[j] – '0';a[index++] = t;}}BigInt operator +(const BigInt &b)const{BigInt res;res.len = max(len,b.len);for(int i = 0;i <= res.len;i++)res.a[i] = 0;for(int i = 0;i < res.len;i++){res.a[i] += ((i < len)?a[i]:0)+((i < b.len)?b.a[i]:0);res.a[i+1] += res.a[i]/mod;res.a[i] %= mod;}if(res.a[res.len] > 0)res.len++;return res;}BigInt operator *(const BigInt &b)const{BigInt res;for(int i = 0; i < len;i++){int up = 0;for(int j = 0;j < b.len;j++){int temp = a[i]*b.a[j] + res.a[i+j] + up;res.a[i+j] = temp%mod;up = temp/mod;}if(up != 0)res.a[i + b.len] = up;}res.len = len + b.len;while(res.a[res.len – 1] == 0 &&res.len > 1)res.len–;return res;}void output(){printf("%d",a[len-1]);for(int i = len-2;i >=0 ;i–)printf("%04d",a[i]);printf("\n");}};BigInt dp[300];int main(){int n;dp[0]=1;dp[1]=1;for(int i=2;i<=250;i++){dp[i]=dp[i-2]+dp[i-2]+dp[i-1];}while(~scanf("%d",&n)){dp[n].output();}return 0;}

自己选择的路,跪着也要把它走完。

POJ 2506 Tiling(高精度+递推)

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