矩阵取数游戏
高精度是套用的一个模板。
#include <algorithm> // max
#include <cassert> // assert
#include <cstdio> // printf,sprintf
#include <cstring> // strlen
#include <iostream> // cin,cout
#include <string> // string类
#include <vector> // vector类
using namespace std;
struct BigInteger {
typedef unsigned long long LL;
static const int BASE = 100000000;
static const int WIDTH = 8;
vector<int> s;
BigInteger& clean(){while(!s.back()&&s.size()>1)s.pop_back(); return *this;}
BigInteger(LL num = 0) {*this = num;}
BigInteger(string s) {*this = s;}
BigInteger& operator = (long long num) {
s.clear();
do {
s.push_back(num % BASE);
num /= BASE;
} while (num > 0);
return *this;
}
BigInteger& operator = (const string& str) {
s.clear();
int x, len = (str.length() - 1) / WIDTH + 1;
for (int i = 0; i < len; i++) {
int end = str.length() - i*WIDTH;
int start = max(0, end - WIDTH);
sscanf(str.substr(start,end-start).c_str(), "%d", &x);
s.push_back(x);
}
return (*this).clean();
}
BigInteger operator + (const BigInteger& b) const {
BigInteger c; c.s.clear();
for (int i = 0, g = 0; ; i++) {
if (g == 0 && i >= s.size() && i >= b.s.size()) break;
int x = g;
if (i < s.size()) x += s[i];
if (i < b.s.size()) x += b.s[i];
c.s.push_back(x % BASE);
g = x / BASE;
}
return c;
}
BigInteger operator - (const BigInteger& b) const {
assert(b <= *this); // 减数不能大于被减数
BigInteger c; c.s.clear();
for (int i = 0, g = 0; ; i++) {
if (g == 0 && i >= s.size() && i >= b.s.size()) break;
int x = s[i] + g;
if (i < b.s.size()) x -= b.s[i];
if (x < 0) {g = -1; x += BASE;} else g = 0;
c.s.push_back(x);
}
return c.clean();
}
BigInteger operator * (const BigInteger& b) const {
int i, j; LL g;
vector<LL> v(s.size()+b.s.size(), 0);
BigInteger c; c.s.clear();
for(i=0;i<s.size();i++) for(j=0;j<b.s.size();j++) v[i+j]+=LL(s[i])*b.s[j];
for (i = 0, g = 0; ; i++) {
if (g ==0 && i >= v.size()) break;
LL x = v[i] + g;
c.s.push_back(x % BASE);
g = x / BASE;
}
return c.clean();
}
BigInteger operator / (const BigInteger& b) const {
assert(b > 0); // 除数必须大于0
BigInteger c = *this; // 商:主要是让c.s和(*this).s的vector一样大
BigInteger m; // 余数:初始化为0
for (int i = s.size()-1; i >= 0; i--) {
m = m*BASE + s[i];
c.s[i] = bsearch(b, m);
m -= b*c.s[i];
}
return c.clean();
}
BigInteger operator % (const BigInteger& b) const { //方法与除法相同
BigInteger c = *this;
BigInteger m;
for (int i = s.size()-1; i >= 0; i--) {
m = m*BASE + s[i];
c.s[i] = bsearch(b, m);
m -= b*c.s[i];
}
return m;
}
// 二分法找出满足bx<=m的最大的x
int bsearch(const BigInteger& b, const BigInteger& m) const{
int L = 0, R = BASE-1, x;
while (1) {
x = (L+R)>>1;
if (b*x<=m) {if (b*(x+1)>m) return x; else L = x;}
else R = x;
}
}
BigInteger& operator += (const BigInteger& b) {*this = *this + b; return *this;}
BigInteger& operator -= (const BigInteger& b) {*this = *this - b; return *this;}
BigInteger& operator *= (const BigInteger& b) {*this = *this * b; return *this;}
BigInteger& operator /= (const BigInteger& b) {*this = *this / b; return *this;}
BigInteger& operator %= (const BigInteger& b) {*this = *this % b; return *this;}
bool operator < (const BigInteger& b) const {
if (s.size() != b.s.size()) return s.size() < b.s.size();
for (int i = s.size()-1; i >= 0; i--)
if (s[i] != b.s[i]) return s[i] < b.s[i];
return false;
}
bool operator >(const BigInteger& b) const{return b < *this;}
bool operator<=(const BigInteger& b) const{return !(b < *this);}
bool operator>=(const BigInteger& b) const{return !(*this < b);}
bool operator!=(const BigInteger& b) const{return b < *this || *this < b;}
bool operator==(const BigInteger& b) const{return !(b < *this) && !(b > *this);}
};
ostream& operator << (ostream& out, const BigInteger& x) {
out << x.s.back();
for (int i = x.s.size()-2; i >= 0; i--) {
char buf[20];
sprintf(buf, "%08d", x.s[i]);
for (int j = 0; j < strlen(buf); j++) out << buf[j];
}
return out;
}
istream& operator >> (istream& in, BigInteger& x) {
string s;
if (!(in >> s)) return in;
x = s;
return in;
}
const int maxn=105;
#define ll BigInteger
int n,m;
int a[maxn];
ll dp[maxn][maxn];
ll ans=0;
int main(){
cin>>n>>m;
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
cin>>a[j];
}
for(int x=0;x<m;x++){
for(int y=0;y<m;y++){
dp[x][y]=0;
}
}
ll mi=1;
for(int j=0;j<m;j++)mi*=2;
for(int j=0;j<m;j++){
dp[j][j]=mi*a[j];
}
for(int len=2;len<=m;len++){
mi/=2;//chu2
for(int j=0;j<=m-len;j++){//遍历每一个大小为len的区间
dp[j][j+len-1]=max(dp[j+1][j+len-1]+mi*a[j],dp[j][j+len-2]+mi*a[j+len-1]);
}
}
ans+=dp[0][m-1];
}
cout<<ans<<endl;
return 0;
}
另外，我看了一下别人的代码。
以下是我喜欢的一个风格。
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int MAXN = 80 + 1;
const int MAXM = 80 + 1;
const int SIZE = 100 + 1;
string add(string s1, string s2) {
int a[SIZE] = { 0 }, b[SIZE] = { 0 }, c[SIZE] = { 0 };
int size_a = s1.size(), size_b = s2.size();
for (int i = 0; i < size_a; i++) a[size_a - i] = s1[i] - '0';
for (int i = 0; i < size_b; i++) b[size_b - i] = s2[i] - '0';
int size_c = 1, x = 0;
while (size_c <= size_a || size_c <= size_b) {
c[size_c] = a[size_c] + b[size_c] + x;
x = c[size_c] / 10;
c[size_c] %= 10;
size_c++;
}
c[size_c] = x;
while (size_c > 1 && c[size_c] == 0) size_c--;
string res;
res.resize(size_c);
for (int i = size_c; i >= 1; i--) res[size_c - i] = c[i] + '0';
return res;
}
string mul(string s1, string s2) {
int a[SIZE] = { 0 }, b[SIZE] = { 0 }, c[SIZE] = { 0 };
int size_a = s1.size(), size_b = s2.size();
for (int i = 0; i < size_a; i++) a[size_a - i] = s1[i] - '0';
for (int i = 0; i < size_b; i++) b[size_b - i] = s2[i] - '0';
for (int i = 1; i <= size_a; i++) {
int x = 0;
for (int j = 1; j <= size_b; j++) {
int now = i + j - 1;
c[now] += a[i] * b[j] + x;
x = c[now] / 10;
c[now] %= 10;
}
c[i + size_b] = x;
}
int size_c = size_a + size_b;
while (size_c > 1 && c[size_c] == 0) size_c--;
string res;
res.resize(size_c);
for (int i = size_c; i >= 1; i--) res[size_c - i] = c[i] + '0';
return res;
}
int n, m;
string a[MAXM], pow2[MAXM], f[MAXN][MAXM], ans;
inline string max(string a, string b) {
if (a.size() > b.size()) return a;
if (a.size() < b.size()) return b;
for (int i = 0; i < a.size(); i++) {
if (a[i] > b[i]) return a;
if (a[i] < b[i]) return b;
}
return a;
}
int main(int argc, char *argv[]) {
cin >> n >> m;
pow2[0] = "1";
for (int i = 1; i <= m; i++) pow2[i] = mul(pow2[i - 1], "2");
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
cin >> a[j];
f[j][j] = mul(pow2[m], a[j]);
}
for (int d = 1; d <= m - 1; d++) {
for (int l = 1; l <= m - d; l++) {
int r = l + d;
f[l][r] = max(add(f[l + 1][r], mul(pow2[m - d], a[l])), add(f[l][r - 1], mul(pow2[m - d], a[r])));
}
}
ans = add(ans, f[1][m]);
}
cout << ans << endl;
return 0;
}
