等比数列二分求和
今天我們學(xué)習(xí)如何有效地求表達(dá)式的值。對(duì)于這個(gè)問題，用二分解決比較好。
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（1）當(dāng)時(shí)，
（2）當(dāng)時(shí)，那么有
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（3）當(dāng)時(shí)，那么有
???
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代碼：
#include <iostream> #include <string.h> #include <stdio.h>using namespace std; const int M = 1000000007; typedef long long LL;LL power(LL a,LL b) {LL ans = 1;a %= M;while(b){if(b & 1){ans = ans * a % M;b--;}b >>= 1;a = a * a % M;}return ans; }LL sum(LL a,LL n) {if(n == 1) return a;LL t = sum(a,n/2);if(n & 1){LL cur = power(a,n/2+1);t = (t + t * cur % M) % M;t = (t + cur) % M;}else{LL cur = power(a,n/2);t = (t + t * cur % M) % M;}return t; }int main() {LL a,n;while(cin>>a>>n)cout<<sum(a,n)<<endl;return 0; }
題目：http://poj.org/problem?id=3233
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題意：矩陣求和
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代碼：
#include <iostream> #include <string.h> #include <stdio.h>using namespace std; const int N = 35;struct Matrix {int m[N][N]; };Matrix I; int n,k,M;Matrix add(Matrix a,Matrix b) {Matrix c;for(int i=0; i<n; i++){for(int j=0; j<n; j++){c.m[i][j] = a.m[i][j] + b.m[i][j];c.m[i][j] %= M;}}return c; }Matrix multi(Matrix a,Matrix b) {Matrix c;for(int i=0; i<n; i++){for(int j=0; j<n; j++){c.m[i][j] = 0;for(int k=0; k<n; k++)c.m[i][j] += a.m[i][k] * b.m[k][j];c.m[i][j] %= M;}}return c; }Matrix power(Matrix A,int n) {Matrix ans = I,p = A;while(n){if(n & 1){ans = multi(ans,p);n--;}n >>= 1;p = multi(p,p);}return ans; }Matrix sum(Matrix A,int k) {if(k == 1) return A;Matrix t = sum(A,k/2);if(k & 1){Matrix cur = power(A,k/2+1);t = add(t,multi(t,cur));t = add(t,cur);}else{Matrix cur = power(A,k/2);t = add(t,multi(t,cur));}return t; }int main() {while(scanf("%d%d%d",&n,&k,&M)!=EOF){Matrix A;for(int i=0; i<n; i++){for(int j=0; j<n; j++){scanf("%d",&A.m[i][j]);A.m[i][j] %= M;I.m[i][j] = (i==j);}}Matrix ans = sum(A,k);for(int i=0; i<n; i++){for(int j=0; j<n; j++)printf("%d ",ans.m[i][j]);puts("");}}return 0; }
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總結(jié)
